Differential Equations

Differential Equations

4 2 2 3sin , ' 2 0, 0y x y y xy x y y x

Definition A differential equation is an equation involving derivatives of an unknown function and possibly the function itself as well as the independent variable.

Example

Definition The order of a differential equation is the highest order of the derivatives of the unknown function appearing in the equation

1st order equations 2nd order equation

sin cosy x y x CExamples 2 3

1 1 26 e 3 e ex x xy x y x C y x C x C

In the simplest cases, equations may be solved by direct integration.

Observe that the set of solutions to the above 1st order equation has 1 parameter, while the solutions to the above 2nd order equation depend on two parameters.

Linear or non-linear

Differential equations are said to be non-linear if any products exist between the dependent variable and its derivatives, or between the derivatives themselves.

xeydx

dy

dx

yd x cos44)( 23

3

Product between two derivatives ---- non-linear

xydx

dycos4 2

Product between the dependent variable themselves ---- non-linear

4

An object is dropped from a height h at t = 0. Determine velocity v and displacement y.

dvg

dt

1v gt C 1 0C v gt

Solution by immediate integrationFREE FALL

dygt

dt

22

1

2y gt C

2 0C 21

2y gt

Separable Differential Equations

A separable differential equation can be expressed as the product of a function of x and a function of y.

dyg x h y

dx

22dy

xydx

Multiply both sides by dx and divide both

sides by y2 to separate the variables.

(Assume y2 is never zero.)

22

dyx dx

y

2 2 y dy x dx

0h y

Example

A separable differential equation can be expressed as the product of a function of x and a function of y.

dyg x h y

dx

22dy

xydx

22

dyx dx

y

2 2 y dy x dx

2 2 y dy x dx 1 2

1 2y C x C

21x C

y

2

1y

x C

2

1y

x C

0h y

Combined constants of integration

Separable Differential Equations

Example

Family of solutions (general solution) of a differential equation

The picture on the right shows some solutions to the above differential equation.

The straight lines

y = x and y = -x

are special solutions. A unique solution curve goes through any point of the plane different from the origin. The special solutions y = x and y = -x go both through the origin.

Cxy

xdxydyy

x

dx

dy

22

Example

Initial conditions

• In many physical problems we need to find the particular solution that satisfies a condition of the form y(x0)=y0. This is called an initial condition, and the problem of finding a solution of the differential equation that satisfies the initial condition is called an initial-value problem.

• Find a solution to y2 = x2 + C satisfying the initial condition y(0) = 2.

22 = 02 + C

C = 4

y2 = x2 + 4

Example

222 1 xdyx y e

dx

2

2

12

1xdy x e dx

y

Separable differential equation

2

2

12

1xdy x e dx

y

2u x

2 du x dx

2

1

1udy e du

y

1

1 2tan uy C e C 21

1 2tan xy C e C 21tan xy e C Combined constants of integration

Example

222 1 xdyx y e

dx

21tan xy e C We now have y as an implicit function

of x.

We can find y as an explicit function of x by taking the tangent of both sides.

21tan tan tan xy e C

2

tan xy e C

A population of living creatures normally increases at a rate that is proportional to the current level of the population. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by:

dyky

dt

Law of natural growth or decay

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

ln y kt Ce e Exponentiate both sides.

C kty e e C kty e e kty Ae

Example of the first order linear ODE – RC circuit

CR uuu

Find the time dependence of the electric current i(t) in the given circuit.

1.u Ri idt

C

Now we take the first derivative of the last equation with respect to time

011

0 iRCdt

dii

Cdt

diR

The general solution is:

tRCKei1

Constant K can be calculated from initial conditions. We know that

R

uKKe

R

u

R

ui RC

0

)0(

tRCe

R

ui

1

particular solution is