Lectures on Differential Equations

PBC Lecture Notes Series: Differential Equations / Dr. Abhijit Kar Gupta 1

Ordinary Differential Equations (Combined Lectures, 1st Ed.)

Lecture Notes prepared by-

Dr. Abhijit Kar Gupta Physics Department, Panskura Banamali College

Panskura R.S., East Midnapore, WB, India, Pin-code: 721152 E-mail: [email protected]

Lecture-1

Books to be consulted:

1. Differential Equations with Applications and Historical Notes - George F. Simmons (Tata McGraw-Hill)

2. Differential and Integral Calculus I & II - N. Piskunov (MIR Pub., Moscow)

3. Introduction to Mathematical Physics - Charlie Harper (PHI) 4. Mathematical Physics - H. K. Dass (S. Chand & Company Ltd.) We start from Definition: A differential equation is one which connects the independent variable x , unknown

function )(xfy = and its derivativesdxdy , 2

2

dxyd ,…… n

n

dxyd .

In general, 0......,,, 2

2

=

dx

yddxdyyxF …………………….. (1)

Example:

02

2

=−+ cdxdyk

dxyd ……………………………………………….(2)

In Physics:

dtdxkmg

dtxdm −=2

2

………………………………………………(3)

A body of mass m is falling under gravity through a medium where it offers a force of resistance proportional to the speed of the body. One has to solve the equation to know the instantaneous position and velocity of the body.

• TYPE of differential equation:


If the function )(xfy = is a function of one independent variable x , then the differential equation is called ORDINARY.

• ORDER of differential equation: The order of a differential equation is the order of the highest derivative which appears in the equation.

For example, the equations (2) and (3) are 2nd order and an equation like 0105 2 =+− xydxdy

is

1st order equation. TO SOLVE:

We have to find out the form of )(xfy = from any given differential equation. We can easily check the correctness of a solution if we put back the relation )(xfy = into the corresponding differential equation and arrive at an identity (both sides equal). Example: Consider the equation for a simple harmonic motion (SHM),

022

2

=+ ydx

yd ω . [an ordinary 2nd order differential equation]

The functions xAy ωsin= and xBy ωcos= will satisfy the equation, where A and B are constants. Therefore, these are two solutions of the SHM equation which we obtain by trial. In general, the function xBxAy ωω cossin += will also satisfy the equation and thus a general solution.

Solutions of First Order Differential Equations

• Equations with separated variables (or separable variables): Let us consider

)()( 21 yfxfdxdy

= ……………………………(1)

To solve the above equation,

dxxfdyyf

)()(

11

2

= …………………………(2)

L.H.S. of equation (2) is a function of y only and the R.H.S. is a function of x alone. Thus the variables x and y are separated. Hence, to get the solution, we integrate on both sides of (2) and obtain,


∫∫ += .)()(

11

2

cdxxfdyyf

The equation (2) can also be written in the following form: 0)()( =+ dyyNdxxM .

Therefore, any 1st order equation of the above form is also having separated variables. Example:

#1. 0=+yx

dxdy

The above can be written in the form .0=+ ydyxdx

∴We get after integrating, cyx=+

22

22

.

#2. Example from Physics: Radioactive Decay If m is the mass of a radioactive substance (Radium) at any time t , then the rate of change of the mass (rate of decay) is as follows:

kmdtdm

−= ,

where k is a constant. To solve we separate the variables t and m .

kdtmdm

−= .

Integrating on both sides, cktm lnln +−=

Or, ktcem −= . Boundary Condition: At the start of the decay process, 0=t , the mass of the radioactive substance was

0mm = . ∴ ktemm −= 0 ⇒ This is Radioactive decay law.

Lecture-2

Homogeneous Function:

Let us have a function ),( yxf . If we replace x by xλ and y by yλ ,

m

t

0m


i.e., xx λ→ and yy λ→ and we get the following identity ),(),( yxfyxf nλλλ =

Then the function ),( yxf is called a Homogeneous function of degree n . Examples: #1. 2),( yxyyxf −= 222),( yxyyxf λλλλ −= = ),()( 222 yxfyxy λλ =− This is a homogeneous function of degree 2.

#2. xy

yxyxf22

),( −=

xyyxyxf 2

222 )(),(λ

λλλ −= =

xyyx 22 − = ),(0 yxfλ

This is a homogeneous function of degree zero.

Homogeneous Equation:

An equation of the type ),( yxfdxdy

= , where ),( yxf is a homogeneous function in x

and y with degree zero is called a homogeneous equation.

# <Example>

22 yxxy

dxdy

−= …………………………..(1)

The right hand side of equation (1) is a homogeneous function of degree zero which can be verified by putting xx λ→ and yy λ→ . Therefore, the given equation is a homogeneous equation. The function can be rearranged to write:

2222 /1/),(

xyxy

yxxyyxf

−=

−= .

Let us substitute xyu /= .

∴ uxy = ⇒ dxduxu

dxdy

+= .

Equation (1) now becomes 21 uu

dxduxu

−=+

⇒ =dxdux u

uu

−− 21

= 2

3

1 uu−

Solution of 1st Order Ordinary Homogeneous differential equation


⇒ x

dxduu

u=

−3

21 [The variables are now separated]

Integrating both sides,

∫∫ =

−

xdxdu

uu11

3

⇒ cxuu

lnlnln21

2 +=−−

⇒ 221)ln(u

cux −=

⇒ 221)ln(u

cy −= [since, uxy = ]

⇒ )ln(2 cyyx −= This is the solution.

# <Example>

13

−−−+

=yxyx

dxdy …………………………(1)

Here it can be checked that the function 13),(

−−−+

=yxyxyxf is not a homogeneous

function because of the constants. If we substitute hxx += 1 and kyy += 1 then

13

11

11

1

1

−−−+−+++

=kyhxkyhx

dxdy =

)1()3(

11

11

−−+−−+++

khyxkhyx = ),( 11 yxf .

),( 11 yxf will be homogeneous if 03 =−+ kh ……………….(i) and 01 =−− kh ……………… (ii) Solving (i) and (ii) we get 2=h , 1=k . Now we have to solve

11

11

11

11

1

1

/1/1xyxy

yxyx

dxdy

−+

=−+

= ……………………….(2)

We substitute 11 / xyu = . ⇒ 11 uxy =

∴ 1

11

1

dxduxu

dxdy

+=

Hence, the equation (1) becomes

Some 1st Order Inhomogeneous differential equations can be reduced to Homogeneous equations.


uu

dxduxu

−+

=+11

11 ⇒

uu

dxdux

−+

=1

1 2

11

⇒ 1

121

1x

dxduuu

=+− [The variables are now separated]


∫∫ =

+−

+ 1

122 11

1x

dxduu

uu

⇒ ( ) cxuu lnln1ln21tan 1

21 +=+−−

⇒ |1|lntan 21

1 ucxu +=−

⇒ ueucx1tan2

1 1−

=+

Putting back 1

1

xyu = , 11

1 /tan21

21

xyeyxc−

=+

Now replacing 21 −=−= xhxx and 11 −=−= ykyy , we get the final solution:

−−−

=−+− 21tan

221

)1()2( xy

eyxc .

Lecture-3

First Order Linear Equations of the following type:

)()( xQyxPdxdy

=+ Linear in the unknown function y and its derivative dxdy .

Method of separation of variables We ask the solution in the form )()( xvxuy = # <Example>

3)1(1

2+=

+− xy

xdxdy …………………………..(1)

General Prescription: • Any 1st order Homogeneous diff. equation can be reduced to an

equation with separated variables and then it can be solved by integration.

• If any 1st order Non-Homogeneous equation can be reduced to Homogeneous one then the previous step can be followed.


Here

+−=

12)(

xxP and ( )31)( += xxQ

Let us put .uvy = ∴ We then have dxduv

dxdvu

dxdy

+= .

Equation (1) reduces to 3)1(

12

+=

+−+ xuv

xdxduv

dxdvu

⇒ 3)1(1

2+=+

+− x

dxduvv

xdxdvu ………………………………(2)

Suppose we choose the function v such that the term

01

2=

+− v

xdxdv .

∴ dxxv

dv

+=

12 ⇒ cxv ln|1|ln2||ln ++= ⇒ 2)1( += xcv .

Putting the above expression of v in (2) we get the following equation in u :

32 )1()1( +=+ xdxduxc

⇒ 2)1(1+= x

cdxdu

Integrating, BxAu ′++′=2

)1( 2

, where A′ and B′ are constants.

∴ The complete solution (complete integral) of the given equation (1) will be of the form 2

4

)1(2

)1(++

+= xBxAy . Solution.

If in the linear equation (1) the functions )(xP and )(xQ are just constants then the solution is easy to find.

baydxdy

=+ , where a and b are constants.

Then baydxdy

+−= The variables are readily separable.

⇒ dxbay

dy=

+− ⇒ 1||ln1 cxbay

a+=+−−

⇒ )( 2caxebay +−=+− [ acc .12 = ]

∴ abcey ax += − [ 2

1 cea

c −−= ]

First Order Nonlinear Equations: Bernoulli’s Equation

Note:


nyxQyxP

dxdy )()( =+

Equation of the above type is called Bernoulli’s equation. Dividing both sides by ny we get,

)()( 1 xQyxPdxdyy nn =+ +−− This is a nonlinear equation.

The above can be converted into a linear equation if we substitute

We have dxdyyn

dxdz n−+−= )1(

∴ )()()1(

1 xQzxPdxdz

n=+

+− ⇒ )()1()()1( xQnzxPn

dxdz

+−=+−+ .

So, the above equation is a linear equation again. Now we substitute uvz = as before and proceed to find solution. # <Example>

33 yxxydxdy

=+ ………………….(1)

Dividing both sides by 3y , we have 323 xxy

dxdyy =+ −− ………………(2)

Substitute 2−= yz to have

dxdyy

dxdz 32 −−= …………………..(3)

Put (3) in (2),

322 xxzdxdz

−=− ……………..(4)

The above is a linear equation where xxP 2)( −= , 32)( xxQ −= . To find the complete integral we substitute uvz = .

∴ dxduv

dxdvu

dxdz

+=

Put the expressions of z and dxdz back in (4),

322 xxuvdxduv

dxdvu −=−+

1+−= nyz


⇒ 322 xdxduvxv

dxdvu −=+

− .

Now v can be chosen such that the expression in the bracket may be zero.

Therefore, 02 =− xvdxdv ⇒ xdx

vdv 2=

Or, 2||ln xv = ⇒ 2xev = .

To find u , we have now 32xdxduv −= .

∴ 322

xdxdue x −=

⇒ 32

2 xedxdu x−−=

⇒ cdxxeu x +−= ∫ − 32

2 Integrating by parts we find:

ceexu xx ++= −− 223 ∴

2

12 xcexuvz ++== ⇒

2

122 xcexy ++=− . Solution.

Lecture-4

Exact Differential Equations

Any 1st order differential equation can be written in the following form:

0),(),( =+ dyyxNdxyxM ……………………….(1) where ),( yxM and ),( yxN are differentiable functions of x and y . We can say, the equation (1) is exact if it can be written in the form: 0),( =yxdf ……………………….(2) Equation (2) can be integrated easily and the complete integral (solution) is Cyxf =),( , where C is an integration constant. # <Example>

0=+yx

dxdy

The above can be written in the form 0=+ ydyxdx


⇒ 0)(21 22 =+ yxd

⇒ 0)( 22 =+ yxd ∴ Cyx =+ 22 Soln.

• Condition for Exactness:

),(),(),( yxdfdyyxNdxyxM =+ = dyyfdx

xf

∂∂

+∂∂

Then

xfM∂∂

= …………………….(A)

yfN∂∂

= ……………………..(B)

Differentiating (A) with respect to y and (B) with respect to x , we obtain

xyf

yM

∂∂∂

=∂∂ 2

yxf

xN

∂∂∂

=∂∂ 2

.

Assuming continuity of the second derivative we can write, The above is a necessary condition for the equation (1) to be exact differential one. To find a solution: From the relation (A) we can write,

∫ += )(),( ydxyxMf φ , after Integrating. Here )(yφ is an integration constant which is a function of y only. Again differentiating the above we get,

)(ydxy

Myf φ′+

∂∂

=∂∂

∫ = ),( yxN [ from the relation (B)]

Then we proceed to find )(yφ . # <Example> To examine if the following equation is exact and then solve:

0324

22

3 =

−+ dy

yxydx

yx …………………………..(1)

xN

yM

∂∂

=∂∂


3

2yxM =

4

22 3y

xyN −=

We have

4

6yx

yM

−=∂∂ and

4

6yx

xN

−=∂∂ (provided 0≠y )

The given equation (1) is thus an exact differential equation. This means .0),( =yxdf To find the complete integral:

Let us take 3

2yx

xfM =∂∂

= .


)(23 ydx

yxf φ+= ∫

∴ )(3

2

yyxf φ+= …………………………..(2)

Differentiating (2) on both sides w.r.t. y and noting that

4

22 3y

xyNyf −

==∂∂

We find

)(34

2

yyx φ′+− = 4

22 3y

xy − .

∴ )(yφ′ = 2

1y

⇒ )(yφ = 11 cy+−

So, we now have from (2),

13

2 11),(cyy

xyxf +−=

Thus the complete integral of the given equation is =),( yxf constant

⇒ Cyy

x=−

13

2

.

Home Work Problems: Check if the following equations are exact. In that case find the solutions,

#1. ( ) 022 2 =++ xydxdyx


#2. 02cos22sin =+ xydxdyx

#3. ( ) 0=−++ xydxdyyx with boundary condition 0)1( =y

#4. ( ) 02 22 =++ xydxdyyx with boundary condition 2)1( =y .

Lecture-5

Integrating Factor

If the equation 0),(),( =+ dyyxNdxyxM ……………….(1) is not an exact differential equation then we can sometimes choose a function ),( yxµ such that after w multiply the original equation by this, the equation is converted into an exact differential one. The function ),( yxµ is called the Integrating factor of the given equation. After multiplying by the integrating factor: 0=+ NdyMdx µµ ……………….(2) Equation (2) will be exact when we have

( ) ( )Nx

My

µµ∂∂

=∂∂

⇒ x

NxN

yM

yM

∂∂

+∂∂

=∂∂

+∂∂ µµµµ

⇒

∂∂

−∂∂

=∂∂

−∂∂

yM

xN

xN

yM µµµ

⇒

∂∂

−∂∂

=∂∂

−∂∂

yM

xN

xN

yM µ

µµ

µ11

⇒

∂∂

−∂∂

=∂

∂−

∂∂

yM

xN

xN

yM µµ lnln …………………(3)

Therefore, the integrating factor ),( yxµ has to satisfy the above relation. • In a special case, if µ is a function of y only then we can write

∂∂

−∂∂

=y

MxN

dydM µln ⇒

∂∂

−∂∂

=y

MxN

Mdyd 1lnµ

Similarly, if µ is a function of x only then we can write

∂∂

−∂∂

=−y

MxN

dxdN µln ⇒

∂∂

−∂∂

−=y

MxN

Ndxd 1lnµ .


# <Example> To solve the following equation: ( ) 02 =−+ xdydxxyy ……………(1) Here, 2xyyM += , xN −=

∴ xyy

M 21+=∂∂ , 1−=

∂∂

xN

Thus xN

yM

∂∂

≠∂∂ . Therefore, the given equation is not exact differential equation.

Here we have

)21(2211 xyxyy

MxN

+−=−−−=

∂∂

−∂∂

Also it is seen that

∂∂

−∂∂

yM

xN

M1 = 2

)1(2xyy

xy++− =

y2

− which is a function of y only.

Thus we conclude that the possible integrating factor may be a function of y only. Therefore,

ydy

d 2ln−=

µ

⇒ yln2ln −=µ , after integrating

⇒ 2

1y

=µ , so this is the choice of integrating factor.

Multiplying the original equation (1) by this integrating factor,

012 =−

+ dy

yxdxx

y……………….(2)

Now it can be verified that 2

1yx

Ny

M−=

∂∂

=∂∂ .

Therefore, the equation (2) is exact differential equation. Equation (2) can be rewritten as

02

2

=

+

xyxd

∴ Cxyx

=+2

2

, C is integration constant.

∴ Cx

xy2

22 +

−= Solution.

[Equation (2) can also be solved formally assuming xfM∂∂

= and yfN∂∂

= etc.]


Special Case: If the 1st order equation can be written in the form:

)()( xQyxPdxdy

=+

Then the Integrating factor can be shown to be # <Example>

xydxdy

=+ 2

Integrating factor: xe2

∴ xxx xeyedxdye 222 2 =+

⇒ ( ) xx xeyedxd 22 =

⇒ dxxeye xx ∫= 22

⇒ .41

21 222 cexeye xxx +−=

Lecture-6

Second-Order Differential Equations

We consider differential equations of the following type:

)(2

2

xfqydxdyp

dxyd

=++

• The equation is LINEAR as this is first degree in y and its derivatives. • The equation has constant coefficients: p and q . • The equation will be called HOMOGENEOUS when 0)( =xf . Otherwise, it is

called non-homogeneous equation.

Solutions for Homogeneous Equation: 02

2

=++ qydxdyp

dxyd

(i) If 1y and 2y are two particular solutions then )( 21 yy + is also a solution. This can be easily verified by putting the solutions back in the equations.

(ii) If 1y is a solution and C is a constant then 1Cy is also a solution.

∫ dxxPe

)(


(iii) If 1y and 2y are two independent solutions then 2211 yCyCy += is a general solution, where 1C and 2C are constants.

Note: To find out the general solution of the Homogeneous equation what we have to do is to find two linearly independent solutions and then the general solution can be constructed by combining the. What do you mean by ‘linearly independent’? If )(1 xy and )(2 xy are two solutions and we have

=)()(

2

1

xyxy )(xφ , [some function of x ]

then the solutions are linearly independent. Otherwise, if

=)()(

2

1

xyxy Constant (independent of x ), the solutions are dependent.

Method of Trial:

Homogeneous equation: 02

2

=++ qydxdyp

dxyd ……………..(1)

Let us look for the particular solutions in the form: kxey = , where k = Constant. Then

kxkedxdy

=

kxekdx

yd 22

2

=

Substituting the above two in equation (1), we get ( ) 02 =++ qpkkekx . Since 0≠kxe , 02 =++ qpkk …………….(2) Equation (2) is called auxiliary equation. The solutions of the quadratic equation:

qppk −+−=42

2

1 , qppk −−−=42

2

2

The following cases are possible:

(I) 1k and 2k are real numbers and 21 kk ≠ (II) 1k and 2k are complex numbers (III) 1k and 2k are real numbers and 21 kk = .


Two unequal real roots 21 kk ≠ We have two independent particular solutions: xkey 1

1 = and xkey 21 =

The complete integral or general solution can be written as xkxk ececy 21

21 += # Example: 02 =−′+′′ yyy The auxiliary equation is 022 =−+ kk Two roots are 11 =k , 22 −=k ∴ The complete integral is xx ececy 2

21−+=

Two complex roots They come in pairs (complex conjugates) βα ik +=1 , βα ik −=2

Here 2p

−=α and 4

2pq −=β .

Two particular solutions are xiey )(

1βα+= and xiey )(

2βα−=

Splitting the Real and Imaginary parts, ( ) xiexexixey xxx ββββ ααα sincossincos1 +=+= ( ) xiexexixey xxx ββββ ααα sincossincos2 −=−=

Now it can be shown that each real and imaginary part can be a particular solution of the original equation: 1y = xe x βα cos (real part) 2y = xe x βα sin (imaginary part) Also we see that ≠= xyy βcot21 constant. Thus we can say we have two independent particular solutions 1y and 2y . Hence, the general solution is 2211 ycycy +=

Case-II

Case-I


= ( )xcxce x ββα sincos 21 + . # Example: 09 =+′′ yy The auxiliary equation is 092 =+k Two roots are ik 31 = , ik 32 −= Here 0=α and 3=β The general solution is xcxcy 3sin3cos 21 += .

Two roots are real and equal kkk == 21 One particular solution is kxey =1 . To find another particular solution independent of the above we take, kxxey =2 . Now we can check that 1y and 2y both satisfy the differential equation and they are linearly independent. ∴ The general solution is kxkx xececy 21 += = )( 21 xccekx + . # Example: 044 =+′−′′ yyy The auxiliary equation is 0442 =+− kk Two roots are 221 == kk The complete integral (general solution) is then zx xececy 2

22

1 += . Lecture-7

Solutions for NonHomogeneous Equation:

)(2

2

xfqydxdyp

dxyd

=++ …………………..(1)

Case-III


The general solution of the nonhomogeneous equation (1) is written as sum of • The general solution of the corresponding homogeneous part y • Particular solution of the nonhomogeneous equation (1), *y

The sum *yyy += is the general solution of the equation (1),

A General Method for finding particular solution of a nonhomogeneous equation Let the right hand side of equation (1) is of the following form:

Where )(xPn is a polynomial of degree n . Case-I The number α is not a root of the auxiliary equation 02 =++ qpkk of the corresponding homogeneous equation:

02

2

=++ qydxdyp

dxyd …………………..(2)

Then we can seek the particular solution in the form x

n exQy α)(* = , ………………………...(3) where )(xQn is a polynomial of degree n . Substituting (3) in (2) and equating the coefficients of the same degrees on both sides, we can determine the unknown coefficients in n

nn xaxaxaaxQ ++++= ...............)( 2210 .

Case-II When the number α is a simple (single) root of the auxiliary equation, we seek the particular solution of the form: x

n exxQy α)(* = . Case-III When the number α is a double root of the auxiliary equation, we seek the particular solution of the form: x

n exQxy α)(2* = . Examples: #1. Find the general solution of the equation xyyy =++ 34 /// …………….(1)

xn exPxf α)()( =


Soln. The corresponding homogeneous equation of the given nonhomogeneous 2nd order differential equation (1) is 034 =+′+′′ yyy ……………………….(2) Considering the trial solution kxey = we get the auxiliary equation: 0342 =++ kk The two roots are 11 −=k , 32 −=k . Therefore, the general solution for the homogeneous equation (2) is xx ececy 3

21−− += .

To find the particular solution of the given nonhomogeneous equation (1), we consider xexxf .0.)( = [ This is of the form xexP α)(1 with 0=α ] Here we seek the particular solution of the form xexQy .0

1* )(=

Thus we take )( 10

* xaay +=

∴ 1/* ay = , 0//* =y .

Substituting the above in (1) we find xxaaa =++ )(34 101 . Equating the coefficients of identical powers of x on both sides, we get 13 1 =a 043 10 =+ aa We obtain

31

1 =a , 94

0 −=a

∴ 94

31* −= xy

Thus the general solution of the given nonhomogeneous equation is *yyy +=

⇒ 94

313

21 −++= −− xececy xx .

• Home Work problem:

Find the general solution of the equation xexyy 32// )1(9 +=+

Ans. xexxxcxcy 3221 81

5271

1813sin3cos

+−++= .


Lecture-8

Nonhomogeneous second order Linear equations with constant coefficients: Example #1

xexyyy )2(67 /// −=+− ………………………….(1) Soln.

Auxiliary equation for the corresponding homogeneous equation 067 /// =+− yyy is 0672 =+− kk ……………………………………..(2)

The two roots are 11 =k and 62 =k . ∴ The general solution of the homogeneous equation is

xx ececy 621 +=

Here, the right side of the given equation (1) is of the form xexP )(1 and the coefficient 1 in the exponential is a simple root of the auxiliary equation (2). Hence, we seek the particular solution in the form xexxQy )(1

* = or xeBAxxy )(* +=

From above we get )2()( 2/* BAxeeBxAxy xx +++=

)24(2)( 2//* BAxeAeeBxAxy xxx ++++= Putting the above expressions in equation (1), we have [ ] xx exeBxAxBAxBxAxABAxBxAx )2()(6)2(7)(72)24()( 222 −=+++−+−++++ ⇒ xx exeABAx )2()2510( −=+−− Equating the coefficients of identical powers of x on both sides,

225

110−=+−

=−AB

A

Thus we obtain 101

−=A , 259

=B .

Therefore, the particular solution is xexxy

+−=

259

101*

Hence, the general solution of the given (Nonhomogeneous) equation is

xxx exxececy

+−++=

259

1016

21 .

Example #2

xyyy cos252 /// =++


The corresponding auxiliary equation is 0522 =++ kk . Two roots are: ik 211 +−= , ik 211 −−= Therefore, the general solution of the corresponding homogeneous equation is )2sin2cos( 21 xcxcey x += − We seek the particular solution of the honhomogeneous equation in the form: xBxAy sincos* += To determine the constant coefficients A and B we find /*y and //*y and put them back in the given equation. We obtain xxBxAxBxAxBxA cos2)sincos(5)cossin(2sincos =+++−+−− Equating the coefficients of xcos and xsin , we get

052252

=+−−=++−

BABABA

Whence,

52

=A , 51

=B

∴ The general solution of the given equation is *yyy +=

xxxcxce x sin51cos

52)2sin2cos( 21 +++= − .

Example from Physics: Vibrations in Mechanical Systems

022

2

=+ xdt

xd ω Simple Harmonic Motion

022

2

=++ xdtdx

dtxd ωγ Damped S.H.M.

)(22

2

tfxdtdx

dtxd

=++ ωγ Damped and forced S.H.M.

Home Work Problems: 1. 0102 /// =++ yyy 2. xyy 2sin24// =+ 3. xeyyy x cos32 /// −=+− 4. xeyyy 2/// 56 =++ 5. xyy 623 /// −=− 6. xexyy 3// .cos9 =− .


Lecture-9

Some More Problems: #1. 2/// 242 xeyyy x +=++ Hints: The particular solution will have the form DCxBxAey x +++= 2* #2. xxyy 2sin9 2// +=+ #3. xyy 2sin24// =+ , with initial values: 0)0( =y , 0)0(/ =y #4. 554 2/// +=++ xyyy , with initial values: 0)0( =y , 0)0(/ =y

Systems of Ordinary Differential Equations In the solution of many problems it is required to find the functions )(1 xy , )(2 xy …etc. of the same argument x where the functions 1y , 2y …satisfy a system of differential equations as in the following:

),,( 2111 yyxf

dxdy

= ………………….(1)

),,( 2122 yyxf

dxdy

= …………………..(2)

Here, we have two coupled differential equations. The task is to find out the solutions

)(1 xy and )(2 xy . Example #1 Integrate the system:

xzydxdy

++= ……………….(1)

xzydxdz 234 +−−= ………….(2)

with the initial conditions 1)0( ==xy , 0)0( ==xz . To find solutions: Differentiate equation (1)

12

2

++=dxdz

dxdy

dxyd …………...(3)

Put back equation (2) into equation (3):

1)234(2

2

++−−+= xzydxdy

dxyd


−−−++−= xy

dxdyxy

dxdy 3124 [From eqn. (1)]

⇒ 1522

2

+=++ xydxdy

dxyd …………………….(4).

Therefore, we have got a 2nd order differential equation with constant coefficients in above. The general solution of equation (4) is 95)( 21 −++= − xexccy x .

Now we have, xydxdyz −−=

146)22( 212 +−−−= − xexccc x . We use the initial conditions to find out the arbitrary constants 1c and 2c . We find 101 =c and 62 =c . Thus we get the following solutions which satisfy the given set of equations and the initial conditions: 95)610( −++= − xexy x 146)1214( +−−−= − xexz x . Example #2 Integrate the system:

zydtdx

+= ……………….(1)

zxdtdy

+= ……………….(2)

yxdtdz

+= ……………….(3)

To find solutions: Differentiating the first equation with respect to t , we find

zyxyxzxdtdz

dtdy

dtxd

++=+++=+= 2)()(2

2

Eliminating the variables y and z , we get

022

2

=−− xdtdx

dtxd .

So we get a 2nd order homogeneous equation, the general solution of which is tt ececx 2

21 += − . From the above we find


tt ececdtdx 2

21 2+−= −

∴ zdtdxy −= = zecec tt −+− − 2

21 2

Now we put the above relation into equation (3) and get

yxdtdz

+=

= ( ) ( ) zecececec tttt −+−++ −− 221

221 2

∴ teczdtdz 2

23=+

The above is a first order differential equation of z . The solution is tt ececz 2

23 += − . Therefore, tttt ececececy 2

232

21 2 +−+−= −− = tt ececc 2

231 )( ++− − . Lecture-10 Physical Interpretation: Let us consider the vibration in a Mechanical System. Example: Oscillation of a spring

Let us consider the spring is loaded with a mass M .

• If k be the spring constant the a force ky− tends to return the load to equilibrium position. [ Restoring Force]

• The motion of the load M is restricted by a force operating in a direction

opposite to that of the motion. The force is proportional to velocity: dtdyλ− .

[Resistance force or Damping force]

M

y


We can write the differential equation of the motion of the load on the spring. By Newton’s 2nd law we have

dtdyky

dtydM λ−−=2

2

.

Thus we have the following homogeneous 2nd order differential equation of the familiar form: 0/// =++ qypyy ………………….(1)

WhereM

p λ= ,

Mkq = .

If the spring-load system is further acted on by an external force )(tf then the corresponding differential equation can be written as )(/// tfqypyy =++ ………………..(2) Equation (1) is called an equation of FREE OSCILLATIONS and equation (2) is called an equation of FORCED OSCILLATIONS.

FREE OSCILLATIONS

0/// =++ qypyy Corresponding auxiliary equation is 02 =++ qpkk . Two roots are:

qppk −+−=42

2

1

qppk −−−=42

2

2

CASE-I

qp>

4

2

(Case of over damping)

Then the roots 1k and 2k are negative numbers. The general solution: tktk ececy 21

21 += ( 01 <k , 02 <k ) It follows that the displacement ( y ) dies down (approaches zero) as ∞→t (long time). This case corresponds to the fat that the forces of resistance are greater than that due to rigidity. There is no oscillation. CASE-II

qp=

4

2

Then the roots are equal.

The general solution is ( ) tpetccy .2/

21−+= .


Here the displacement )(y approaches zero as ∞→t (long time limit) but comparatively slower than that of case-II. CASE-III

0=p (No resistance, No damping) Equation (1) takes the form 0// =+ qyy . The general solution is: tctcy ωω sincos 21 += , where q=ω . If we write 01 sinφAc = and 02 cosφAc = , the solution becomes )sin( 0φω += tAy . This is a solution of Simple Harmonic motion. CASE-IV (Damped Oscillations)

0≠p , qp<

4

2

In this case the roots of the auxiliary equation are complex numbers: βα ik +=1 , βα ik −=2

Where 2p

−=α < 0, 4/2pq −=β .

The general solution has the form: ( )tctcey t ββα sincos 21 += Or )sin( 0φβα += tAey t . The solution is oscillatory but damped. Oscillation is given by the sin-term. The amplitude of oscillation falls off exponentially (finally goes down to zero) given by teα -term.

The envelope of the peaks falls off exponentially.


Lecture-11 Method of Variation of Parameters: [NOTE: This method is due to Lagrange which is particularly useful to solve 2nd order nonhomogeneous equations with variable coefficients.]

Let us consider the general equation of the form: )()()( xfyxqyxpy =+′+′′ .

Assume the following form of the solution: 221 yvvyy += , (1) where 1y and 2y are two independent solutions of the corresponding homogeneous equations. The coefficients 1v and 2v are two unknown functions. The function )(xf is considered to be continuous in some range. Therefore, we can write

0)()( 111 =+′+′′ yxqyxpy ………………….(2) 0)()( 222 =+′+′′ yxqyxpy ...……………..…(3) Now from (1), we can have 22221111 yvyvyvyvy ′+′+′+′=′ ( ) ( )22112211 yvyvyvyv ′+′+′+′= . (4) To simplify the calculation of y ′′ , we choose 1v and 2v so that 02211 =′+′ yvyv . (5) The final solution corresponding to the given equation will be subject to the condition we choose. The equation (4) now becomes 2211 yvyvy ′+′=′ (6) Now differentiating equation (5) once again with respect to x , we obtain 22221111 yvyvyvyvy ′′+′′+′′+′′=′′ . (7) Substituting equations (1), (6) and (7) into the original equation, we obtain

)()()()()( 2211221122221111 xfyvxqyvxqyvxpyvxpyvyvyvyv =++′+′+′′+′′+′′+′′ ⇒ ( ) ( ) )()()()()( 221122221111 xfyvyvyxqyxpyvyxqyxpyv =′′+′′++′+′′++′+′′ .


The two expressions in parentheses in the last equation vanish as they are essentially the solutions of the homogeneous equation. Hence we now have

)(2211 xfyvyv =′′+′′ . (8)

Solving equations (5) and (8) for 1v′ and 2v′ , we obtain

∆−=

∆

′=′

)()(0

22

2

1xfyyxf

y

v

and

∆=

∆

′=′

)()(0

11

1

2xfyxfy

y

v ,

where

( )212121

21 yyyyyyyy

′−′=′′

=∆ .

The determinant ∆ is called the Wronskian ),( 21 yyW of 1y and 2y , The quantities 1v and 2v can be found out from the above by integration. Therefore, we obtain the following solution:

dxyyWyxfydx

yyWyxfyy ∫∫ +−=

),()(

),()(

21

12

21

21 .

The above is the general solution of the given equation. This method is useful when the two independent solutions 1y and 2y can not be obtained easily. Example #1 To solve: xyy sec=+′′ . [Because of the trigonometric function on the right hand side it is difficult to find out the particular integral by the methods described earlier. However, the present method will be very useful as we see.] From the homogeneous equation, we find two independent solutions to be

xy sin1 = and xy cos2 = .

∴ The Wronskian is 21

2121 ),(

yyyy

yyW′′

=∆= = 1sincos

cossin−=

− xxxx

.


Therefore, the particular integral is

∫ ∫−= xdxxxxdxxxy p sin.seccoscos.secsin

= dxxxxdxx ∫∫ −−

cossincossin

= )log(coscossin xxxx +− . The complete solution is thus

).log(coscossincossin xxxxxBxAy +−+= Example #2

To solve: xxyyxyx ln222 =+′−′′ ; ( )0≠x Soln: The standard form for the equation is

xxy

xy

xy ln22

2 =+′−′′ ,

where x

xp 2)( −=

2

2)(x

xq =

xxxf ln)( =

Let us take two independent solutions of the corresponding homogeneous equation by investigation * . We have xy =1 and 2

2 xy = . Hence we have

22

21

2121 21),( x

xxx

yyyy

yyW ==′′

=∆= .

Thus the general solution is

dxx

xxx

xdxx

xxx

xy ∫∫

+

−= 22

2

2 lnln

dxx

xxdxxxx ∫∫ +−= 2

2 lnln

( )

+−−+

+−= 2

21

2 1lnln21 c

xxxxcxx

( ) 221

2 1lnln21 xcxcxxx +−

++−= .

* Actually, the solutions can be found out by a standard method and the given equation is of the type of Cauchy-Euler equation of order 2.


Cauchy-Euler Equation: If we have an equation of the following type

02 =+′+′′ byyaxyx , we search for the solutions of the algebraic form

mxxy =)( . In the above example, we have the homogeneous equation :

0222 =+′−′′ yyxyx To solve, we look for the solution of the form, mxxy =)( . Thus we shall have 1−=′ mmxy and 2)1( −−=′′ mxmmy . Hence the corresponding auxiliary equation becomes

0232 =+− mm . The two roots are 11 =m and 22 =m . Therefore, we have two independent solutions

xy =1 and 22 xy = .

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Lectures on Differential Equations

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