Maths Transform MethodsTopic 02 : ODEs

Lectures 1012 : Introduction to Ordinary Differential Equations (ODEs)

Dr Kieran MurphyThis module and thesenotes were developedby Dr Pardaig Kirwanwith only minor modi-fications on my part.

Credits:

Department of Computing and Mathematics,Waterford Institute of Technology.

(kmurphy@wit.ie)

Autumn Semester, 2014

OutlineTypes of ODEs

Solution of ODEs using integration

Application to electrical circuits

1 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Many models of engineering systems involve the rate of change of a quantity.

There is thus a need to incorporate derivatives into the mathematical model.

These mathematical models are examples of differential equations.

Accompanying the differential equation will be one or more conditions that letus obtain a unique solution to a particular problem.

Often we solve the differential equation first to obtain a general solution; thenwe apply the conditions to obtain the unique solution.

It is important to know which conditions must be specified in order to obtain aunique solution.

2 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When a hot liquid is placed in a cooler environment, experimental observationshows that its temperature decreases to approximately that of its surroundings.

A typical graph of the temperature of the liquid plotted against time is shownbelow.

.

.T

Ts

t

3 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

After an initially rapid decrease the temperature changes progressively less rapidlyand eventually the curve appears to flatten out.

Newtons law of coolingThe rate of cooling of liquid is proportional to the difference between itstemperature and the temperature of its environment (the ambient temperature).

To convert this into mathematics, lett be the time elapsed (in seconds, s),T the temperature of the liquid ( C),andT0 the temperature of the liquid at the start (t = 0).

The temperature of the surroundings is denoted by Ts.

ExampleWrite down the mathematical equation which is equivalent to Newtons law ofcooling and state the accompanying condition.

4 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

After an initially rapid decrease the temperature changes progressively less rapidlyand eventually the curve appears to flatten out.

Newtons law of coolingThe rate of cooling of liquid is proportional to the difference between itstemperature and the temperature of its environment (the ambient temperature).

To convert this into mathematics, lett be the time elapsed (in seconds, s),T the temperature of the liquid ( C),andT0 the temperature of the liquid at the start (t = 0).

The temperature of the surroundings is denoted by Ts.

ExampleWrite down the mathematical equation which is equivalent to Newtons law ofcooling and state the accompanying condition.

4 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

After an initially rapid decrease the temperature changes progressively less rapidlyand eventually the curve appears to flatten out.

Newtons law of coolingThe rate of cooling of liquid is proportional to the difference between itstemperature and the temperature of its environment (the ambient temperature).

To convert this into mathematics, lett be the time elapsed (in seconds, s),T the temperature of the liquid ( C),andT0 the temperature of the liquid at the start (t = 0).

The temperature of the surroundings is denoted by Ts.

ExampleWrite down the mathematical equation which is equivalent to Newtons law ofcooling and state the accompanying condition.

4 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

After an initially rapid decrease the temperature changes progressively less rapidlyand eventually the curve appears to flatten out.

Newtons law of coolingThe rate of cooling of liquid is proportional to the difference between itstemperature and the temperature of its environment (the ambient temperature).

To convert this into mathematics, lett be the time elapsed (in seconds, s),T the temperature of the liquid ( C),andT0 the temperature of the liquid at the start (t = 0).

The temperature of the surroundings is denoted by Ts.

ExampleWrite down the mathematical equation which is equivalent to Newtons law ofcooling and state the accompanying condition.

4 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

In the above example we call t the independent variable and T the dependentvariable.Since the condition is given at t = 0 we refer to it as an initial condition.The solution of the above differential equation which satisfies the initialcondition is T = Ts + (T0 Ts)ekt

DefinitionA differential equation is a relationship between the value of a function y(t) andthe values of its derivatives, which is true for all valid values of t. y(t) is called asolution of the differential equation.

ExampleShow that y = Ae2t satisfies the differential equation

dydt

= 2t

5 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

In the above example we call t the independent variable and T the dependentvariable.Since the condition is given at t = 0 we refer to it as an initial condition.The solution of the above differential equation which satisfies the initialcondition is T = Ts + (T0 Ts)ekt

DefinitionA differential equation is a relationship between the value of a function y(t) andthe values of its derivatives, which is true for all valid values of t. y(t) is called asolution of the differential equation.

ExampleShow that y = Ae2t satisfies the differential equation

dydt

= 2t

5 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleShow that y = (A + Bt)e2t is a solution of the differential equation

d2ydt2

+ 4dydt

+ 4y = 0

ExampleShow that y = A sin(3t) + B cos(3t) is a solution of the differential equation

d2ydt2

+ 9y = 0

ExampleDetermine a differential equation satisfied by

y = A cosh(2t) + B sinh(2t)

6 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleShow that y = (A + Bt)e2t is a solution of the differential equation

d2ydt2

+ 4dydt

+ 4y = 0

ExampleShow that y = A sin(3t) + B cos(3t) is a solution of the differential equation

d2ydt2

+ 9y = 0

ExampleDetermine a differential equation satisfied by

y = A cosh(2t) + B sinh(2t) 6 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

We have seen that an expression including one arbitrary constant required onedifferentiation to obtain a differential equation which eliminated the arbitraryconstant.

Where two constants were present, two differentiations were required.

Is the converse true? For example,

would a differential equation involving dydt as the only derivative have a generalsolution with one arbitrary constant and

would a differential equation which had d2ydt2 as the highest derivative produce a

general solution with two arbitrary constants?

The answer is, usually, yes.

7 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIntegrate twice the differential equation

d2ydt2

= 8(6t t2)

To determine values for A and B we need two additional pieces of information.This might take the form of the value of y or its derivative for given values of t.This would yield a unique solution to the differential equation.

ExampleDetermine the unique solution of the differential equation

dydt

= 4t3

which satisfies the condition y(1) = 12.

8 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIntegrate twice the differential equation

d2ydt2

= 8(6t t2)

To determine values for A and B we need two additional pieces of information.This might take the form of the value of y or its derivative for given values of t.This would yield a unique solution to the differential equation.

ExampleDetermine the unique solution of the differential equation

dydt

= 4t3

which satisfies the condition y(1) = 12.

8 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIntegrate twice the differential equation

d2ydt2

= 8(6t t2)

To determine values for A and B we need two additional pieces of information.This might take the form of the value of y or its derivative for given values of t.This would yield a unique solution to the differential equation.

ExampleDetermine the unique solution of the differential equation

dydt

= 4t3

which satisfies the condition y(1) = 12.

8 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

It is worth noting that different additional conditions give rise to different solutionsto a given differential equation.

Example

Solve the differential equationd2ydt2

= 4t subject to the conditions

(a) y(0) = 2 and y(1) = 6

(b) y(0) = 2 anddydt

= 3 at t = 0.

9 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

It is worth noting that different additional conditions give rise to different solutionsto a given differential equation.

Example

Solve the differential equationd2ydt2

= 4t subject to the conditions

(a) y(0) = 2 and y(1) = 6

(b) y(0) = 2 anddydt

= 3 at t = 0.

9 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When solving differential equations (either analytically or numerically) it isimportant to be able to recognise the various kinds that can arise. We therefore needto introduce some terminology which will help us to distinguish one kind ofdifferential equation from another.

An ordinary differential equation (ODE) is any relation between a function ofa single variable and its derivatives. (All differential equations studied in thisworkbook are ordinary.)

The order of a differential equation is the order of the highest derivative in theequation.

A differential equation is linear if the dependent variable and its derivativesoccur to the first power only and if there are no products involving thedependent variable or its derivatives.

We will concern ourselves only with linear ordinary differential equations of orderat most 2.

10 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When solving differential equations (either analytically or numerically) it isimportant to be able to recognise the various kinds that can arise. We therefore needto introduce some terminology which will help us to distinguish one kind ofdifferential equation from another.

An ordinary differential equation (ODE) is any relation between a function ofa single variable and its derivatives. (All differential equations studied in thisworkbook are ordinary.)

The order of a differential equation is the order of the highest derivative in theequation.

A differential equation is linear if the dependent variable and its derivativesoccur to the first power only and if there are no products involving thedependent variable or its derivatives.

We will concern ourselves only with linear ordinary differential equations of orderat most 2.

10 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When solving differential equations (either analytically or numerically) it isimportant to be able to recognise the various kinds that can arise. We therefore needto introduce some terminology which will help us to distinguish one kind ofdifferential equation from another.

An ordinary differential equation (ODE) is any relation between a function ofa single variable and its derivatives. (All differential equations studied in thisworkbook are ordinary.)

The order of a differential equation is the order of the highest derivative in theequation.

A differential equation is linear if the dependent variable and its derivativesoccur to the first power only and if there are no products involving thedependent variable or its derivatives.

We will concern ourselves only with linear ordinary differential equations of orderat most 2.

10 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When solving differential equations (either analytically or numerically) it isimportant to be able to recognise the various kinds that can arise. We therefore needto introduce some terminology which will help us to distinguish one kind ofdifferential equation from another.

An ordinary differential equation (ODE) is any relation between a function ofa single variable and its derivatives. (All differential equations studied in thisworkbook are ordinary.)

The order of a differential equation is the order of the highest derivative in theequation.

A differential equation is linear if the dependent variable and its derivativesoccur to the first power only and if there are no products involving thedependent variable or its derivatives.

We will concern ourselves only with linear ordinary differential equations of orderat most 2.

10 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

When solving differential equations (either analytically or numerically) it isimportant to be able to recognise the various kinds that can arise. We therefore needto introduce some terminology which will help us to distinguish one kind ofdifferential equation from another.

An ordinary differential equation (ODE) is any relation between a function ofa single variable and its derivatives. (All differential equations studied in thisworkbook are ordinary.)

The order of a differential equation is the order of the highest derivative in theequation.

A differential equation is linear if the dependent variable and its derivativesoccur to the first power only and if there are no products involving thedependent variable or its derivatives.

We will concern ourselves only with linear ordinary differential equations of orderat most 2.

10 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleClassify the differential equations, specifying the order and type (linear/non-linear).

(a)d2ydt2 dy

dt= t2

(b)d2ydt2

=

(dydt

)3+ 3t

(c)dxdt t = x2

(d)dydt

+ cos(y) = 0

(e)dydt

+ y2 = 4

11 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

General electrical circuits involving inductors, capacitors and resistors can bemodelled by second order linear ordinary differential equations.

We will initially assume that the voltage applied to the circuit is alwaysconstant.

We will deal with the case of varying voltage supplies in the LaplaceTransforms section.

We need to recall some basic laws of electricity.

Ohms LawThe voltage across the resistor is iR where i is the current flowing in the circuit andR is the (constant) resistance.

Inductor Voltage DropThe voltage across the inductance is L didt where L is the constant inductance.

12 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Capacitor Voltage DropThe voltage across the capacitor is qC where q is the charge on the capacitor, and Cis the capacitance.

DefinitionCurrent is the rate of change of charge. If q(t) represents the charge on a capacitorthen

i(t) =dqdt

NoteWe shall assume that the resistance, capacitance and inductance shall remainconstant for each problem. This means that our differential equations will only haveconstant coefficients of the objective function and its derivatives.

13 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Capacitor Voltage DropThe voltage across the capacitor is qC where q is the charge on the capacitor, and Cis the capacitance.

DefinitionCurrent is the rate of change of charge. If q(t) represents the charge on a capacitorthen

i(t) =dqdt

NoteWe shall assume that the resistance, capacitance and inductance shall remainconstant for each problem. This means that our differential equations will only haveconstant coefficients of the objective function and its derivatives.

13 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

Kirchhoffs law of voltagesThe applied voltage is the sum of the other voltages in the circuit.

ExampleIn this RL circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R L

E

Write down a differential equation for the current i and state the initial condition.

14 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIn this RC circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C

E

Write down a differential equation for the current i and state the initial condition.

15 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIn this LC circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

L C

E

Write down a differential equation for the current i and state the initial condition.

16 of 1

Introduction to Differential Equations

Introduction to Differential Equations Lecture 10

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Write down a differential equation for the current i and state the initial condition.

17 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Both RC and LR circuits give rise to first-order linear differential equations.We have previously seen that equations of this type have solutions that involvethe exponential function.

ExampleShow that y = 3e4t satisfies the differential equation

dydt

+ 4y = 0

ExampleShow that y = 2 + 4e2t satisfies the differential equation

dydt 2y = 4

18 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Both RC and LR circuits give rise to first-order linear differential equations.We have previously seen that equations of this type have solutions that involvethe exponential function.

ExampleShow that y = 3e4t satisfies the differential equation

dydt

+ 4y = 0

ExampleShow that y = 2 + 4e2t satisfies the differential equation

dydt 2y = 4

18 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Both RC and LR circuits give rise to first-order linear differential equations.We have previously seen that equations of this type have solutions that involvethe exponential function.

ExampleShow that y = 3e4t satisfies the differential equation

dydt

+ 4y = 0

ExampleShow that y = 2 + 4e2t satisfies the differential equation

dydt 2y = 4

18 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Both RC and LR circuits give rise to first-order linear differential equations.We have previously seen that equations of this type have solutions that involvethe exponential function.

ExampleShow that y = 3e4t satisfies the differential equation

dydt

+ 4y = 0

ExampleShow that y = 2 + 4e2t satisfies the differential equation

dydt 2y = 4

18 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Both RC and LR circuits give rise to first-order linear differential equations.We have previously seen that equations of this type have solutions that involvethe exponential function.

ExampleShow that y = 3e4t satisfies the differential equation

dydt

+ 4y = 0

ExampleShow that y = 2 + 4e2t satisfies the differential equation

dydt 2y = 4

18 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Solution to First-order Differential EquationsThe general situation is that

d(Aekt)dt

= A(kekt)

= k[Aekt]

i.e. the function y = Aekt satisfies the differential equation

dydt

= ky

To be more precise about the solution we need more information such as the valueof y or dydt for a given value of t. This gives us an initial value problem.

19 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Solution to First-order Differential EquationsThe general situation is that

d(Aekt)dt

= A(kekt) = k[Aekt]

i.e. the function y = Aekt satisfies the differential equation

dydt

= ky

To be more precise about the solution we need more information such as the valueof y or dydt for a given value of t. This gives us an initial value problem.

19 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Solution to First-order Differential EquationsThe general situation is that

d(Aekt)dt

= A(kekt) = k[Aekt]

i.e. the function y = Aekt satisfies the differential equation

dydt

= ky

To be more precise about the solution we need more information such as the valueof y or dydt for a given value of t. This gives us an initial value problem.

19 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Solution to First-order Differential EquationsThe general situation is that

d(Aekt)dt

= A(kekt) = k[Aekt]

i.e. the function y = Aekt satisfies the differential equation

dydt

= ky

To be more precise about the solution we need more information such as the valueof y or dydt for a given value of t. This gives us an initial value problem.

19 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSolve the following differential equation by inspection

2dx(t)dt

+ 5x(t) = 0 where x(0) = 6

The above are examples of the type

dydt

+ ky = 0

homogeneous

20 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSolve the following differential equation by inspection

2dx(t)dt

+ 5x(t) = 0 where x(0) = 6

The above are examples of the type

dydt

+ ky = 0 homogeneous

20 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Suppose that we have a non-zero constant, A, on the right side of the equation thenwe have an example of the type

dydt

+ ky = A

non-homogeneous

This is easily solved by making the substitution

x(t) = y(t) Ak

ExampleSolve

(a)dydt

+ 7y = 14

(b) 2dydt

+ 5y = 15

21 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Suppose that we have a non-zero constant, A, on the right side of the equation thenwe have an example of the type

dydt

+ ky = A

non-homogeneous

This is easily solved by making the substitution

x(t) = y(t) Ak

ExampleSolve

(a)dydt

+ 7y = 14

(b) 2dydt

+ 5y = 15

21 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Suppose that we have a non-zero constant, A, on the right side of the equation thenwe have an example of the type

dydt

+ ky = A

non-homogeneous

This is easily solved by making the substitution

x(t) = y(t) Ak

ExampleSolve

(a)dydt

+ 7y = 14

(b) 2dydt

+ 5y = 15

21 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Suppose that we have a non-zero constant, A, on the right side of the equation thenwe have an example of the type

dydt

+ ky = A non-homogeneous

This is easily solved by making the substitution

x(t) = y(t) Ak

ExampleSolve

(a)dydt

+ 7y = 14

(b) 2dydt

+ 5y = 15

21 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

Suppose that we have a non-zero constant, A, on the right side of the equation thenwe have an example of the type

dydt

+ ky = A non-homogeneous

This is easily solved by making the substitution

x(t) = y(t) Ak

ExampleSolve

(a)dydt

+ 7y = 14

(b) 2dydt

+ 5y = 15

21 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

The solution of the previous problem

y(t) = Ae52 + 3

is called a general solution of the differential equation as all possible solutionsto the equation are of this type.

To uniquely determine a value for A we need more information such as thevalue of y(t) or dydt for a given value of t.

This gives us an initial value problem.

ExampleSolve the initial value problems

(a)dxdt

+ 34x = 0 x(0) = 2

(b)dxdt

+ 3x = 12 x(0) = 2

22 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

The solution of the previous problem

y(t) = Ae52 + 3

is called a general solution of the differential equation as all possible solutionsto the equation are of this type.

To uniquely determine a value for A we need more information such as thevalue of y(t) or dydt for a given value of t.

This gives us an initial value problem.

ExampleSolve the initial value problems

(a)dxdt

+ 34x = 0 x(0) = 2

(b)dxdt

+ 3x = 12 x(0) = 2

22 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

The solution of the previous problem

y(t) = Ae52 + 3

is called a general solution of the differential equation as all possible solutionsto the equation are of this type.

To uniquely determine a value for A we need more information such as thevalue of y(t) or dydt for a given value of t.

This gives us an initial value problem.

ExampleSolve the initial value problems

(a)dxdt

+ 34x = 0 x(0) = 2

(b)dxdt

+ 3x = 12 x(0) = 2

22 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

The solution of the previous problem

y(t) = Ae52 + 3

is called a general solution of the differential equation as all possible solutionsto the equation are of this type.

To uniquely determine a value for A we need more information such as thevalue of y(t) or dydt for a given value of t.

This gives us an initial value problem.

ExampleSolve the initial value problems

(a)dxdt

+ 34x = 0 x(0) = 2

(b)dxdt

+ 3x = 12 x(0) = 2

22 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graph of

v(t) = 3 + 4et

Determine a differential equation which has v(t) as a solution by analysing the

relationship betweendvdt

and v(t).

We note that the above graph is non-zero for negative values of t.Since t usually denotes time we often want the function to be zero for t < 0.This usually involves the Heaviside step function.

23 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graph of

v(t) = 3 + 4et

Determine a differential equation which has v(t) as a solution by analysing the

relationship betweendvdt

and v(t).

We note that the above graph is non-zero for negative values of t.Since t usually denotes time we often want the function to be zero for t < 0.This usually involves the Heaviside step function.

23 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graph of

v(t) = 3 + 4et

Determine a differential equation which has v(t) as a solution by analysing the

relationship betweendvdt

and v(t).

We note that the above graph is non-zero for negative values of t.

Since t usually denotes time we often want the function to be zero for t < 0.This usually involves the Heaviside step function.

23 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graph of

v(t) = 3 + 4et

Determine a differential equation which has v(t) as a solution by analysing the

relationship betweendvdt

and v(t).

We note that the above graph is non-zero for negative values of t.Since t usually denotes time we often want the function to be zero for t < 0.

This usually involves the Heaviside step function.

23 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graph of

v(t) = 3 + 4et

Determine a differential equation which has v(t) as a solution by analysing the

relationship betweendvdt

and v(t).

We note that the above graph is non-zero for negative values of t.Since t usually denotes time we often want the function to be zero for t < 0.This usually involves the Heaviside step function.

23 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleSketch the graphs of each of the following:(a) e7tU(t)(b) 4e7tU(t)(c) 4e7tU(t)(d) (2 4e7t)U(t)

24 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleIn this RL circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R L

E

Determine an expression for the current i if the inductance is L = 0.5 henry and theresistance is 10, given that(a) E = 0 volts.(b) E = 12 volts.Assume that there is no initial current in the circuit.

25 of 1

Introduction to Differential Equations Linear First-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 11

ExampleIn this RC circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C

E

Determine an expression for the current i if the capacitance is C = 0.01 Farad andthe resistance is 100, given that(a) E = 0 volts.(b) E = 12 volts.Assume that there is no initial charge on the capacitor.

26 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

In this Section we start to learn how to solve second order differentialequations of a particular type: those that are linear and have constantcoefficients.Such equations are used widely in the modelling of physical phenomena, forexample, in the analysis of vibrating systems and the analysis of electricalcircuits.The general form of such an equation is:

ad2ydt2

+ bdydt

+ cy = f (t)

where a, b, c are constants. This equation is said to be homogeneous iff (t) = 0 for all t.We will initially study such equations where b = 0, i.e. there is no firstderivative term in the differential equation.Our analysis will break into two parts;

where c = k2 is a negative numberwhere c = 2 is a positive number.

27 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Recall that differentiating cosh(kt) and sinh(kt) twice is the same as multiplicationby k2. Thus x(t) = Acosh(kt) + Bsinh(kt) is the general solution of the differentialequation

d2x(t)dt2

k2x(t) = 0

ExampleSolve the differential equations

1d2x(t)dt2

49x(t) = 0

2d2x(t)dt2

4x(t) = 16 - a non-homogeneous problem

NoteProblems of this type do not arise in LC circuits. Otherwise we would have anegative value for the inductance or the capacitance.

28 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Recall that differentiating cosh(kt) and sinh(kt) twice is the same as multiplicationby k2. Thus x(t) = Acosh(kt) + Bsinh(kt) is the general solution of the differentialequation

d2x(t)dt2

k2x(t) = 0

ExampleSolve the differential equations

1d2x(t)dt2

49x(t) = 0

2d2x(t)dt2

4x(t) = 16 - a non-homogeneous problem

NoteProblems of this type do not arise in LC circuits. Otherwise we would have anegative value for the inductance or the capacitance.

28 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Recall that differentiating cos(t) and sin(t) twice is the same as multiplicationby 2. Thus

x(t) = A cos(t) + B sin(t)

is the general solution of the second order, linear, homogeneous differentialequation

d2x(t)dt2

+ 2x(t) = 0

ExampleSolve the differential equations

1d2x(t)dt2

+ 16pi2x(t) = 0

2d2x(t)dt2

+ 8x(t) = 16 - a non-homogeneous problem

29 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Recall that differentiating cos(t) and sin(t) twice is the same as multiplicationby 2. Thus

x(t) = A cos(t) + B sin(t)

is the general solution of the second order, linear, homogeneous differentialequation

d2x(t)dt2

+ 2x(t) = 0

ExampleSolve the differential equations

1d2x(t)dt2

+ 16pi2x(t) = 0

2d2x(t)dt2

+ 8x(t) = 16 - a non-homogeneous problem

29 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LC circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

L C

E

Determine an expression for the current i if the capacitance is C = 0.02 Farad andthe inductance is 0.1 henry, given that(a) E = 0 volts.(b) E = 12 volts.Assume that there is no initial charge on the capacitor and no initial current.

30 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleDetermine, by inspection, the function x(t) which is the general solution of thedifferential equation

md2x(t)dt2

+ kx(t) = 0

if m = 2 and k = 2 104. Hence determine the particular solution that satisfies theinitial conditions x(0) = 0 and x(0) = 1.

31 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

General LCR circuits with no forcing function (external voltage supply) can bemodelled by second-order linear homogeneous differential equations such as

ad2ydt2

+ bdydt

+ cy = 0

To assist in solving problems of this type we make the following observation.

NoteIf y1(t) and y2(t) are two solutions of a second-order linear homogeneousdifferential equation, such that y1(t) is not a multiple of y2(t), then the generalsolution of the differential equation is

y(t) = Ay1(t) + By2(t)

where A,B are constants.

32 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

General LCR circuits with no forcing function (external voltage supply) can bemodelled by second-order linear homogeneous differential equations such as

ad2ydt2

+ bdydt

+ cy = 0

To assist in solving problems of this type we make the following observation.

NoteIf y1(t) and y2(t) are two solutions of a second-order linear homogeneousdifferential equation, such that y1(t) is not a multiple of y2(t), then the generalsolution of the differential equation is

y(t) = Ay1(t) + By2(t)

where A,B are constants.

32 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleShow that y1 = e3t and y2 = e2t both satisfy the second-order linear homogeneousdifferential equation:

d2ydt2 dy

dt 6y = 0

Write down the general solution of this equation.

ExampleDetermine values of k such that y = ekt is a solution of

d2ydt2

+ 3dydt

+ 2y = 0

Hence state the general solution.

33 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleShow that y1 = e3t and y2 = e2t both satisfy the second-order linear homogeneousdifferential equation:

d2ydt2 dy

dt 6y = 0

Write down the general solution of this equation.

ExampleDetermine values of k such that y = ekt is a solution of

d2ydt2

+ 3dydt

+ 2y = 0

Hence state the general solution.

33 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

The solution of a second-order linear homogeneous differential equation is closelyrelated to solving a quadratic equation. This is made explicit as follows.

DefinitionThe auxiliary equation of the second-order linear homogeneous differentialequation

ad2ydt2

+ bdydt

+ cy = 0

isak2 + bk + c = 0

where y = ekt.

34 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

The solution of a second-order linear homogeneous differential equation is closelyrelated to solving a quadratic equation. This is made explicit as follows.

DefinitionThe auxiliary equation of the second-order linear homogeneous differentialequation

ad2ydt2

+ bdydt

+ cy = 0

isak2 + bk + c = 0

where y = ekt.

34 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleDetermine the auxiliary equations of the following:

(a)d2ydt2

+dydt

+ y = 0

(b) 4d2ydt2

+ 7y = 0

(c) 2d2ydt2

+ 7dydt 3y = 0

(d)d2ydt2

+dydt

= 0

35 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Solving the auxiliary equation gives the values of k which we need todetermine the general solution of the differential equation.

Clearly the nature of the roots will depend on the values of a, b and c.

The solutions of the auxiliary equation can always be determined by thequadratic formula, i.e.

k =bb2 4ac

2a

We will deal with the solutions of the quadratic equation based on the value ofthe discriminant in three cases:

Case 1: b2 > 4ac

Case 2: b2 = 4ac

Case 3: b2 < 4ac

36 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 1: b2 > 4acIf b2 > 4ac then the quadratic formula will yield two real and distinct values for k.The two values of k thus obtained, k1 and k2, will allow us to write down twoindependent solutions:

y1(t) = ek1t and y2(t) = ek2t

and so the general solution of the differential equation will be:

y(t) = Aek1t + Bek2t

ExampleDetermine the general solution of

d2ydt2 7dy

dt+ 12y = 0

37 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 1: b2 > 4acIf b2 > 4ac then the quadratic formula will yield two real and distinct values for k.The two values of k thus obtained, k1 and k2, will allow us to write down twoindependent solutions:

y1(t) = ek1t and y2(t) = ek2t

and so the general solution of the differential equation will be:

y(t) = Aek1t + Bek2t

ExampleDetermine the general solution of

d2ydt2 7dy

dt+ 12y = 0

37 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 0.1 henry, theresistance is 14 and C = 0.25 farad, given that E = 0 volts. Assume that theinitial charge on the capacitor is 0.4 coloumb and that there is no initial current.Sketch the graphs of q(t) and i(t).

Over-damping

If R2 >4LC

then we say that the system is over-damped.

38 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 0.1 henry, theresistance is 14 and C = 0.25 farad, given that E = 0 volts. Assume that theinitial charge on the capacitor is 0.4 coloumb and that there is no initial current.Sketch the graphs of q(t) and i(t).

Over-damping

If R2 >4LC

then we say that the system is over-damped.

38 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 2: b2 = 4acIf b2 = 4ac then the quadratic formula will yield a unique real value for k. Thevalue of k thus obtained will allow us to write down two independent solutions:

y1(t) = ekt and y2(t) = tekt

and so the general solution of the differential equation will be:y(t) = Aekt + Btekt

ExampleDetermine the auxiliary equation of the differential equation

d2xdt2

+ 8dxdt

+ 16x = 0

(a) Show that e4t is a solution.(b) Show that te4t is another solution.(c) Determine the particular solution if

x(0) = 2 and x(0) = 0

39 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 2: b2 = 4acIf b2 = 4ac then the quadratic formula will yield a unique real value for k. Thevalue of k thus obtained will allow us to write down two independent solutions:

y1(t) = ekt and y2(t) = tekt

and so the general solution of the differential equation will be:y(t) = Aekt + Btekt

ExampleDetermine the auxiliary equation of the differential equation

d2xdt2

+ 8dxdt

+ 16x = 0

(a) Show that e4t is a solution.(b) Show that te4t is another solution.(c) Determine the particular solution if

x(0) = 2 and x(0) = 039 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 0.1 henry, theresistance is 8 and C = 1160 farad, given that E = 0 volts. Assume that the initialcharge on the capacitor is 0.2 coloumb and that there is no initial current. Sketchthe graphs of q(t) and i(t).

Critical-damping

If R2 =4LC

then we say that the system is critically-damped.

40 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 0.1 henry, theresistance is 8 and C = 1160 farad, given that E = 0 volts. Assume that the initialcharge on the capacitor is 0.2 coloumb and that there is no initial current. Sketchthe graphs of q(t) and i(t).

Critical-damping

If R2 =4LC

then we say that the system is critically-damped.

40 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

In the final case we will encounter the square root of negative numbers. This bringsus into the realm of complex numbers. Consequently we need to recall someimportant facts.

DefinitionA complex number is an ordered pair of real numbers. We usually denote it as

z = (x, y)

and describe x as the real part of z and y as the imaginary part of z. This is madeexplicit as follows:

x = Re(z) y = Im(z)

We usually denote the set of complex numbers as C and the set of real numbers asR. We have that

R Csince every real number x can be identified with the complex number (x, 0).

41 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

DefinitionWe define

j = (0, 1)

This has the property thatj2 = 1

In some scientific texts we have that i is written instead of j, so that i2 = 1.

Consequently, every complex number can be written as

z = x + jy

Eulers Formula

ei = cos() + j sin()

42 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

DefinitionWe define

j = (0, 1)

This has the property thatj2 = 1

In some scientific texts we have that i is written instead of j, so that i2 = 1.

Consequently, every complex number can be written as

z = x + jy

Eulers Formula

ei = cos() + j sin()

42 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

DefinitionWe define

j = (0, 1)

This has the property thatj2 = 1

In some scientific texts we have that i is written instead of j, so that i2 = 1.

Consequently, every complex number can be written as

z = x + jy

Eulers Formula

ei = cos() + j sin()

42 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 2: b2 < 4acIf b2 < 4ac then the quadratic formula will yield two distinct complex values for k.These values will be complex conjugates and can be denoted by k1 = a + jb andk2 = a jb. Hence

y1(t) = e(a+jb)t and y2(t) = e(ajb)t

and so the general solution of the differential equation will be:

y(t) = Ce(a+jb)t + De(ajb)t

ExampleShow that the general solution of the differential equation in this case will be:

y(t) = eat(A cos(bt) + B sin(bt))

where A,B are real constants.

43 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

Case 2: b2 < 4acIf b2 < 4ac then the quadratic formula will yield two distinct complex values for k.These values will be complex conjugates and can be denoted by k1 = a + jb andk2 = a jb. Hence

y1(t) = e(a+jb)t and y2(t) = e(ajb)t

and so the general solution of the differential equation will be:

y(t) = Ce(a+jb)t + De(ajb)t

ExampleShow that the general solution of the differential equation in this case will be:

y(t) = eat(A cos(bt) + B sin(bt))

where A,B are real constants. 43 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleDetermine the general solution of the differential equation

d2ydt2

+ 16y = 0

ExampleDetermine the general solution of the differential equation

d2ydt2

+ 2dydt

+ 4y = 0

44 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleDetermine the general solution of the differential equation

d2ydt2

+ 16y = 0

ExampleDetermine the general solution of the differential equation

d2ydt2

+ 2dydt

+ 4y = 0

44 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 1 henry, theresistance is 0.2 and C = 0.05 farad, given that E = 0 volts. Assume that theinitial charge on the capacitor is 0.4 coloumb and that there is no initial current.Sketch the graphs of q(t) and i(t).

Over-dampingIf R2 < 4LC then we say that the system is under-damped. If R = 0 then we say thatthe system is undamped. The motion that ensues is called simple harmonicmotion.

45 of 1

Introduction to Differential Equations Linear Second-Order Homogeneous Differential Equations

Introduction to Differential Equations Lecture 12

ExampleIn this LCR circuit the switch is closed at t = 0 and a constant voltage E is applied.

.

.

R C L

E

Determine an expression for the current i if the inductance is L = 1 henry, theresistance is 0.2 and C = 0.05 farad, given that E = 0 volts. Assume that theinitial charge on the capacitor is 0.4 coloumb and that there is no initial current.Sketch the graphs of q(t) and i(t).

Over-dampingIf R2 < 4LC then we say that the system is under-damped. If R = 0 then we say thatthe system is undamped. The motion that ensues is called simple harmonicmotion.

45 of 1