Lectures On Differential Equations 3qa3j

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

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 dny dy d 2 y ,…… . function y = f (x) and its derivatives , dx dx 2 dx n   dy d 2 y In general, F  x, y, , 2 ......  = 0 …………………….. (1) dx dx   Example:

d2y dy +k − c = 0 ……………………………………………….(2) 2 dx dx

In Physics:

d 2x dx = mg − k ………………………………………………(3) 2 dt dt 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. m

•

TYPE of differential equation:

1

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

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If the function y = f (x) 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 1st order equation.

dy − 5 xy 2 + 10 = 0 is dx

TO SOLVE:

We have to find out the form of y = f (x) from any given differential equation. We can easily check the correctness of a solution if we put back the relation y = f (x) into the corresponding differential equation and arrive at an identity (both sides equal). Example:

Consider the equation for a simple harmonic motion (SHM), d2y + ω 2 y = 0 . [an ordinary 2nd order differential equation] 2 dx The functions y = A sin ωx and y = B cos ωx 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 y = A sin ωx + B cos ωx 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 dy = f 1 ( x) f 2 ( y ) ……………………………(1) dx To solve the above equation, 1 dy = f1 ( x)dx …………………………(2) f 2 ( y) 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,

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

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1 dy = ∫ f1 ( x)dx + c. 2 ( y) The equation (2) can also be written in the following form: M ( x)dx + N ( y )dy = 0 . st Therefore, any 1 order equation of the above form is also having separated variables.

∫f

Example:

dy x + =0 dx y The above can be written in the form xdx + ydy = 0. #1.

x2 y2 + =c. 2 2

∴ We get after integrating,

#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: dm = − km , dt where k is a constant. To solve we separate the variables t and m . dm = − kdt . m Integrating on both sides,

ln m = − kt + ln c Or, m = ce − kt .

Boundary Condition: At the start of the decay process, t = 0 , the mass of the radioactive substance was m = m0 .

∴ m = m0 e − kt

⇒ This is Radioactive decay law. m0 m t

Lecture-2 Homogeneous Function: Let us have a function f ( x, y ) . If we replace x by λx and y by λy ,

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

i.e., x → λx and y → λy and we get the following identity f ( λ x , λ y ) = λ n f ( x, y ) Then the function f ( x, y ) is called a Homogeneous function of degree n . Examples: #1. f ( x, y ) = xy − y 2

f (λx, λy ) = λ2 xy − λ2 y 2 = λ2 ( xy − y 2 ) = λ2 f ( x, y ) This is a homogeneous function of degree 2. x2 − y2 #2. f ( x, y ) = xy x2 − y2 λ2 ( x 2 − y 2 ) = = λ 0 f ( x, y ) f ( λx, λy ) = 2 xy λ xy This is a homogeneous function of degree zero. Homogeneous Equation: dy = f ( x, y ) , where f ( x, y ) is a homogeneous function in x An equation of the type dx and y with degree zero is called a homogeneous equation. Solution of 1st Order Ordinary Homogeneous differential equation

# <Example>

dy xy = 2 …………………………..(1) dx x − y 2 The right hand side of equation (1) is a homogeneous function of degree zero which can be verified by putting x → λx and y → λy . Therefore, the given equation is a homogeneous equation. The function can be rearranged to write: xy y/x f ( x, y ) = 2 = . 2 x −y 1− y2 / x2 Let us substitute u = y / x . du dy ∴ y = ux ⇒ =u+x . dx dx du u Equation (1) now becomes u + x = dx 1 − u 2 du u3 u ⇒ x = u = − dx 1 − u 2 1− u2

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

1− u2 dx du = 3 x u

⇒

[The variables are now separated]

Integrating both sides,

 1

∫  u ⇒

−

3

1 dx − du = ∫ u x

1 − ln u = ln x + ln c 2u 2

1 2u 2 1 ln(cy) = − 2 2u x = y − 2 ln(cy )

⇒

ln(cux) = −

⇒ ⇒

[since, y = ux ]

This is the solution.

Some 1st Order Inhomogeneous differential equations can be reduced to Homogeneous equations. # <Example>

dy x + y − 3 = …………………………(1) dx x − y − 1

Here it can be checked that the function f ( x, y ) =

x+ y −3 is not a homogeneous x − y −1

function because of the constants. If we substitute x = x1 + h and y = y1 + k then dy1 x1 + h + y1 + k − 3 x + y1 + (h + k − 3) = = 1 = f ( x1 , y1 ) . dx1 x1 + h − y1 − k − 1 x1 − y1 + (h − k − 1) f ( x1 , y1 ) will be homogeneous if h + k − 3 = 0 ……………….(i) and h − k − 1 = 0 ……………… (ii) Solving (i) and (ii) we get h = 2 , k = 1 . Now we have to solve dy1 x1 + y1 1 + y1 / x1 = = ……………………….(2) dx1 x1 − y1 1 − y1 / x1 u = y1 / x1 . We substitute ⇒ y1 = ux1 dy du ∴ 1 = u + x1 dx1 dx1 Hence, the equation (1) becomes

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

du 1 + u du 1 + u 2 = = ⇒ x1 dx1 1 − u dx1 1 − u dx 1− u ⇒ du = 1 2 x1 1+ u Integrating both sides, dx1 u   1 ∫ 1 + u 2 − 1 + u 2 du = ∫ x1 1 ⇒ tan −1 u − ln 1 + u 2 = ln x1 + ln c 2

6

u + x1

(

[The variables are now separated]

)

⇒ tan −1 u = ln | cx1 1 + u 2 | −1

⇒

cx1 1 + u 2 = e tan u −1 y Putting back u = 1 , c x12 + y12 = e tan y1 / x1 x1 Now replacing x1 = x − h = x − 2 and y1 = y − k = y − 1 , we get the final solution: c ( x − 2) + ( y − 1) = e 2

2

 y −1  tan −1    x−2 

.

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.

Lecture-3 First Order Linear Equations of the following type:

dy + P ( x) y = Q ( x) dx

Linear in the unknown function y and its derivative

Method of separation of variables We ask the solution in the form y = u ( x)v( x) # <Example>

dy  2  3 −  y = ( x + 1) …………………………..(1) dx  x + 1 

dy . dx

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

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 2  3 Here P ( x) = −  and Q( x) = (x + 1)  x + 1

Let us put y = uv.

∴ We then have

du dy dv . =u +v dx dx dx

Equation (1) reduces to dv du  2  3 u +v − uv = ( x + 1) dx dx  x + 1  2  du  dv ⇒ u − v + v = ( x + 1) 3 ………………………………(2) dx  dx x + 1  Suppose we choose the function v such that the term dv  2  − v = 0 . dx  x + 1  dv  2  2 ∴ = dx ⇒ ln | v |= 2 ln | x + 1 | + ln c ⇒ v = c( x + 1) . v  x +1 Putting the above expression of v in (2) we get the following equation in u : du c( x + 1) 2 = ( x + 1) 3 dx du 1 ⇒ = ( x + 1) 2 dx c ( x + 1) 2 Integrating, u = A′ + B ′ , where A′ and B ′ are constants. 2 ∴ The complete solution (complete integral) of the given equation (1) will be of the form ( x + 1) 4 y=A + B( x + 1) 2 . Solution. 2 Note:

If in the linear equation (1) the functions P( x) and Q( x) are just constants then the solution is easy to find. dy + ay = b , where a and b are constants. dx dy = − ay + b The variables are readily separable. Then dx 1 dy ⇒ = dx ⇒ − ln | −ay + b |= x + c1 − ay + b a − ( ax + c2 ) ⇒ − ay + b = e [ c 2 = c1 .a ] b 1 ∴ y = ce − ax + [ c = − e − c2 ] a a First Order Nonlinear Equations: Bernoulli’s Equation

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

dy + P ( x) y = Q ( x) y n dx Equation of the above type is called Bernoulli’s equation. Dividing both sides by y n we get, dy y −n + P( x) y − n +1 = Q( x) This is a nonlinear equation. dx The above can be converted into a linear equation if we substitute

z = y − n +1 dy dz = (−n + 1) y − n dx dx 1 dz dz ∴ + P( x) z = Q( x) ⇒ + (−n + 1) P( x) z = (−n + 1)Q( x) . (−n + 1) dx dx So, the above equation is a linear equation again. Now we substitute z = uv as before and proceed to find solution.

We have

# <Example>

dy + xy = x 3 y 3 ………………….(1) dx Dividing both sides by y 3 , we have dy y −3 + xy − 2 = x 3 ………………(2) dx Substitute z = y −2 to have dz dy = −2 y −3 …………………..(3) dx dx Put (3) in (2), dz − 2 xz = −2 x 3 ……………..(4) dx The above is a linear equation where P( x) = −2 x , Q( x) = −2 x 3 . To find the complete integral we substitute z = uv . dz dv du =u +v ∴ dx dx dx dz back in (4), Put the expressions of z and dx dv du u +v − 2 xuv = −2 x 3 dx dx

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

du  dv  ⇒ u − 2 xv  + v = −2 x 3 . dx  dx  Now v can be chosen such that the expression in the bracket may be zero. dv dv − 2 xv = 0 ⇒ = 2 xdx dx v 2 Or, ln | v |= x 2 ⇒ v = e x . Therefore,

To find u , we have now v ∴ ex

2

du = −2 x 3 . dx

du = −2 x 3 dx

2 du = −2e − x x 3 dx 2 ⇒ u = −2 ∫ e − x x 3 dx + c

⇒

Integrating by parts we find: 2 2 u = x 3e − x + e − x + c 2 ∴ z = uv = x 2 + 1 + ce x 2 ⇒ y −2 = x 2 + 1 + ce x . Solution. Lecture-4 Exact Differential Equations

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

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

dy x + =0 dx y The above can be written in the form xdx + ydy = 0

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

1 d (x2 + y 2 ) = 0 2 ⇒ d (x 2 + y 2 ) = 0 ∴ x2 + y2 = C Soln. ⇒

•

Condition for Exactness:

M ( x, y )dx + N ( x, y )dy = df ( x, y ) =

∂f ∂f dx + dy ∂y ∂x

Then

∂f …………………….(A) ∂x ∂f N= ……………………..(B) ∂y Differentiating (A) with respect to y and (B) with respect to x , we obtain

M =

∂M ∂ 2 f = ∂y ∂y∂x ∂N ∂ 2 f . = ∂x ∂x∂y Assuming continuity of the second derivative we can write,

∂M ∂N = ∂y ∂x

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, f = ∫ M ( x, y )dx + φ ( y ) , after Integrating.

Here φ ( y ) is an integration constant which is a function of y only. Again differentiating the above we get, ∂f ∂M =∫ dx + φ ′( y ) = N ( x, y ) [ from the relation (B)] ∂y ∂y Then we proceed to find φ ( y ) . # <Example>

To examine if the following equation is exact and then solve:  y 2 − 3x 2  2x dy = 0 …………………………..(1)  dx + y3 y4  

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

M =

2x y3

y 2 − 3x 2 y4 We have 6x ∂M = − 4 and ∂y y 6x ∂N =− 4 (provided y ≠ 0 ) ∂x y The given equation (1) is thus an exact differential equation. This means df ( x, y ) = 0. To find the complete integral: ∂f 2 x = . Let us take M = ∂x y 3 Integrating both sides, 2x f = ∫ 3 dx + φ ( y ) y N=

x2 f = 3 + φ ( y ) …………………………..(2) ∴ y Differentiating (2) on both sides w.r.t. y and noting that y 2 − 3x 2 ∂f =N= ∂y y4 We find

y 2 − 3x 2 3x 2 ′ + φ ( y ) = . y4 y4 1 1 ∴ φ ′( y ) = 2 ⇒ φ ( y ) = − + c1 y y So, we now have from (2), x2 1 1 f ( x, y ) = 3 − + y c1 y Thus the complete integral of the given equation is f ( x, y ) = constant −

x2 1 ⇒ 3 − =C. y y Home Work Problems: Check if the following equations are exact. In that case find the solutions,

(

#1. 2 + x 2

) dy + 2 xy = 0 dx

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

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dy + 2 y cos 2 x = 0 dx dy #3. (x + y ) + y − x = 0 with boundary condition y (1) = 0 dx dy #4. (2 + x 2 y ) + xy 2 = 0 with boundary condition y (1) = 2 . dx #2. sin 2 x

Lecture-5 Integrating Factor

If the equation

M ( x, y )dx + N ( x, y )dy = 0 ……………….(1) is not an exact differential equation then we can sometimes choose a function µ ( x, y ) such that after w multiply the original equation by this, the equation is converted into an exact differential one. The function µ ( x, y ) is called the Integrating factor of the given equation. After multiplying by the integrating factor: µMdx + µNdy = 0 ……………….(2) Equation (2) will be exact when we have ∂ (µM ) = ∂ (µN ) ∂y ∂x ∂N ∂µ ∂M ∂µ ⇒ µ +M =µ +N ∂x ∂y ∂y ∂x  ∂N ∂M  ∂µ ∂µ  −N = µ  − ⇒ M ∂y ∂x ∂y   ∂x 1 ∂µ 1 ∂µ  ∂N ∂M   −N = − µ ∂y µ ∂x  ∂x ∂y  ∂ ln µ ∂ ln µ  ∂N ∂M   …………………(3) −N =  − M ⇒ ∂x ∂y  ∂y  ∂x Therefore, the integrating factor µ ( x, y ) has to satisfy the above relation. • In a special case, if µ is a function of y only then we can write

⇒

M

d ln µ  ∂N ∂M  d ln µ 1  ∂N ∂M    ⇒  =  − = − ∂y  ∂y  dy dy M  ∂x  ∂x Similarly, if µ is a function of x only then we can write M

−N

d ln µ  ∂N ∂M =  − ∂y dx  ∂x

 1  ∂N ∂M d ln µ  ⇒ = −  − ∂y dx N  ∂x 

  . 

PBC Lecture Notes Series: Differential Equations / Dr. Abhijit Kar Gupta # <Example>

To solve the following equation: y + xy 2 dx − xdy = 0 ……………(1) Here, M = y + xy 2 , N = − x ∂M ∂N ∴ = 1 + 2 xy , = −1 ∂y ∂x ∂M ∂N Thus ≠ . Therefore, the given equation is not exact differential equation. ∂y ∂x Here we have  ∂N ∂M    = −1 − 1 − 2 xy = −2(1 + 2 xy ) − ∂y   ∂x Also it is seen that

(

)

− 2(1 + xy ) 1  ∂N ∂M  2   = − = − which is a function of y only. 2 ∂y  M  ∂x y y + xy Thus we conclude that the possible integrating factor may be a function of y only. Therefore, d ln µ 2 =− dy y ln µ = −2 ln y , after integrating ⇒ 1 ⇒ µ = 2 , so this is the choice of integrating factor. y Multiplying the original equation (1) by this integrating factor, 1  x  + x dx − 2 dy = 0 ……………….(2) y y  1 ∂M ∂N Now it can be verified that = =− 2 . ∂y ∂x y Therefore, the equation (2) is exact differential equation. Equation (2) can be rewritten as  x x2  d  +  = 0 y 2 

∴ ∴

x x2 + = C , C is integration constant. y 2 2x y=− 2 Solution. x + 2C

[Equation (2) can also be solved formally assuming M =

∂f ∂f and N = etc.] ∂x ∂y

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

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Special Case:

If the 1st order equation can be written in the form: dy + P ( x) y = Q ( x) dx Then the Integrating factor can be shown to be

e∫

P ( x ) dx

# <Example>

dy + 2y = x dx Integrating factor: e 2 x

∴ e2x

dy + 2e 2 x y = xe 2 x dx

d (ye 2 x ) = xe 2 x dx ⇒ ye 2 x = ∫ xe 2 x dx ⇒

⇒ ye 2 x =

1 2x 1 2x xe − e + c. 2 4

Lecture-6 Second-Order Differential Equations

We consider differential equations of the following type: d2y dy + p + qy = f ( x) 2 dx dx

• • •

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 f ( x) = 0 . Otherwise, it is called non-homogeneous equation.

d2y dy + p + qy = 0 2 dx dx If y1 and y 2 are two particular solutions then ( y1 + y 2 ) is also a solution. This

Solutions for Homogeneous Equation: (i)

can be easily verified by putting the solutions back in the equations. (ii)

If y1 is a solution and C is a constant then Cy1 is also a solution.

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

(iii)

15

If y1 and y 2 are two independent solutions then y = C1 y1 + C 2 y 2 is a general solution, where C1 and C 2 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 y1 ( x) and y 2 ( x) are two solutions and we have y1 ( x) = φ ( x) , [some function of x ] y 2 ( x) then the solutions are linearly independent. Otherwise, if y1 ( x) = Constant (independent of x ), the solutions are dependent. y 2 ( x) Method of Trial:

dy d2y + p + qy = 0 ……………..(1) 2 dx dx Let us look for the particular solutions in the form: y = e kx , where k = Constant. Then dy = ke kx dx d2y = k 2 e kx 2 dx Substituting the above two in equation (1), we get e kx (k 2 + pk + q ) = 0 . Since e kx ≠ 0 , k 2 + pk + q = 0 …………….(2) Equation (2) is called auxiliary equation. The solutions of the quadratic equation: Homogeneous equation:

k1 = −

p + 2

p2 p − q , k2 = − − 4 2

The following cases are possible: (I) k1 and k 2 are real numbers and k1 ≠ k 2 (II) k1 and k 2 are complex numbers (III) k1 and k 2 are real numbers and k1 = k 2 .

p2 −q 4

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

Case-I

16

Two unequal real roots k1 ≠ k2

We have two independent particular solutions: y1 = e k1x and y1 = e k 2 x The complete integral or general solution can be written as y = c1e k1 x + c2e k 2 x # Example:

y ′′ + y ′ − 2 y = 0 The auxiliary equation is k2 + k −2 = 0 Two roots are k1 = 1 , k 2 = −2 ∴ The complete integral is y = c1e x + c 2 e −2 x

Case-II

Two complex roots

They come in pairs (complex conjugates) k 1 = α + iβ , k 2 = α − iβ

p p2 and β = q − . 2 4 Two particular solutions are y1 = e (α +iβ ) x and y 2 = e (α −iβ ) x Splitting the Real and Imaginary parts, y1 = eαx (cos β x + i sin βx ) = eαx cos β x + ie αx sin βx y 2 = eαx (cos βx − i sin β x ) = e αx cos β x − ie αx sin βx Here α = −

Now it can be shown that each real and imaginary part can be a particular solution of the original equation: y1 = eαx cos βx (real part)

y 2 = eαx sin β x

(imaginary part)

Also we see that y1 y 2 = cot βx ≠ constant.

Thus we can say we have two independent particular solutions y1 and y 2 . Hence, the general solution is y = c1 y1 + c 2 y 2

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

= eαx (c1 cos βx + c 2 sin βx ) . # Example:

y ′′ + 9 y = 0 The auxiliary equation is k2 +9 = 0 Two roots are k1 = 3i , k 2 = −3i α = 0 and β = 3 Here The general solution is y = c1 cos 3 x + c 2 sin 3 x .

Case-III

Two roots are real and equal k1 = k 2 = k

One particular solution is y1 = e kx . To find another particular solution independent of the above we take, y2 = xekx . Now we can check that y1 and y2 both satisfy the differential equation and they are linearly independent. ∴ The general solution is y = c1e kx + c2 xekx = e kx (c1 + c2 x) . # Example:

y′′ − 4 y′ + 4 y = 0 The auxiliary equation is k 2 − 4k + 4 = 0 Two roots are k1 = k 2 = 2 The complete integral (general solution) is then y = c1e 2 x + c2 xe2 z . Lecture-7 Solutions for NonHomogeneous Equation:

d2y dy + p + qy = f ( x) …………………..(1) 2 dx dx

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

18

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 y = y + y * 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:

f ( x) = Pn ( x)eαx Where Pn ( x) is a polynomial of degree n .

Case-I The number α is not a root of the auxiliary equation k 2 + pk + q = 0 of the corresponding homogeneous equation: dy d2y + p + qy = 0 …………………..(2) 2 dx dx Then we can seek the particular solution in the form y * = Qn ( x)eαx , ………………………...(3) where Qn ( x) 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 Qn ( x) = a 0 + a1 x + a 2 x 2 + ............... + a n x n . Case-II When the number α is a simple (single) root of the auxiliary equation, we seek the particular solution of the form: y * = xQn ( x)eαx . Case-III When the number α is a double root of the auxiliary equation, we seek the particular solution of the form: y * = x 2 Q n ( x ) e αx . Examples: #1. Find the general solution of the equation y // + 4 y / + 3 y = x …………….(1)

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

Soln. The corresponding homogeneous equation of the given nonhomogeneous 2nd order differential equation (1) is y ′′ + 4 y ′ + 3 y = 0 ……………………….(2) Considering the trial solution y = e kx we get the auxiliary equation: k 2 + 4k + 3 = 0

The two roots are k1 = −1 , k 2 = −3 . Therefore, the general solution for the homogeneous equation (2) is y = c1e − x + c 2 e −3 x . To find the particular solution of the given nonhomogeneous equation (1), we consider f ( x) = x.e 0. x [ This is of the form P1 ( x)e αx with α = 0 ] Here we seek the particular solution of the form y * = Q1 ( x)e 0. x Thus we take y * = (a 0 + a1 x) /

//

∴ y * = a1 , y* = 0 . Substituting the above in (1) we find 4a1 + 3(a 0 + a1 x) = x . Equating the coefficients of identical powers of x on both sides, we get 3a1 = 1 3a 0 + 4a1 = 0 We obtain 1 4 a1 = , a 0 = − 3 9 1 4 ∴ y* = x − 3 9 Thus the general solution of the given nonhomogeneous equation is y = y + y* 1 4 ⇒ y = c1e − x + c 2 e −3 x + x − . 3 9 •

Home Work problem:

Find the general solution of the equation y // + 9 y = ( x 2 + 1)e 3 x 1 5 1 Ans. y = c1 cos 3x + c 2 sin 3x +  x 2 − x + e 3 x . 27 81   18

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Lecture-8 Nonhomogeneous second order Linear equations with constant coefficients:

Example #1 y // − 7 y / + 6 y = ( x − 2)e x ………………………….(1) Soln. Auxiliary equation for the corresponding homogeneous equation y // − 7 y / + 6 y = 0 is k 2 − 7k + 6 = 0 ……………………………………..(2) The two roots are k1 = 1 and k 2 = 6 . ∴ The general solution of the homogeneous equation is y = c1e x + c 2 e 6 x

Here, the right side of the given equation (1) is of the form P1 ( x)e x 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 y * = xQ1 ( x)e x or y * = x( Ax + B)e x From above we get

/

y * = ( Ax 2 + Bx)e x + e x (2 Ax + B ) //

y * = ( Ax 2 + Bx)e x + 2 Ae x + e x (4 Ax + 2 B) Putting the above expressions in equation (1), we have

[( Ax

]

+ Bx) + (4 Ax + 2 B) + 2 A − 7( Ax 2 + Bx) − 7(2 Ax + B) + 6( Ax 2 + Bx) e x = ( x − 2)e x ⇒ (−10 Ax − 5B + 2 A)e x = ( x − 2)e x 2

Equating the coefficients of identical powers of x on both sides, − 10 A = 1 − 5 B + 2 A = −2 1 9 . Thus we obtain A = − , B = 10 25 9   1 Therefore, the particular solution is y * = x − x + e x 25   10 Hence, the general solution of the given (Nonhomogeneous) equation is

9   1 y = c1e x + c 2 e 6 x + x − x + e x . 25   10 Example #2 y // + 2 y / + 5 y = 2 cos x

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The corresponding auxiliary equation is k 2 + 2k + 5 = 0 . Two roots are: k1 = −1 + 2i , k1 = −1 − 2i Therefore, the general solution of the corresponding homogeneous equation is y = e − x (c1 cos 2 x + c 2 sin 2 x) We seek the particular solution of the honhomogeneous equation in the form: y * = A cos x + B sin x To determine the constant coefficients A and B we find y * / and y * // and put them back in the given equation. We obtain − A cos x − B sin x + 2(− A sin x + B cos x) + 5( A cos x + B sin x) = 2 cos x Equating the coefficients of cos x and sin x , we get − A + 2B + 5 A = 2 − B − 2 A + 5B = 0 Whence, 2 1 A= , B= 5 5 ∴ The general solution of the given equation is y = y + y*

2 1 = e − x (c1 cos 2 x + c 2 sin 2 x) + cos x + sin x . 5 5 Example from Physics: Vibrations in Mechanical Systems

d 2x +ω2x = 0 2 dt d 2x dx +γ +ω2x = 0 2 dt dt d 2x dx +γ + ω 2 x = f (t ) 2 dt dt Home Work Problems: 1. y // + 2 y / + 10 y = 0 2. y // + 4 y = 2 sin 2 x 3. y // − 2 y / + 3 y = e − x cos x 4. y // + 6 y / + 5 y = e 2 x 5. y // − 3 y / = 2 − 6 x 6. y // − 9 y = cos x.e 3 x .

Simple Harmonic Motion Damped S.H.M. Damped and forced S.H.M.

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Lecture-9 Some More Problems: #1. y // + y / + 2 y = 4e x + 2 x 2 Hints:

The particular solution will have the form y * = Ae x + Bx 2 + Cx + D #2. y // + 9 y = x 2 + sin 2 x #3. y // + 4 y = 2 sin 2 x , with initial values: y (0) = 0 , y / (0) = 0 #4. y // + 4 y / + 5 y = x 2 + 5 , with initial values: y (0) = 0 , y / (0) = 0 Systems of Ordinary Differential Equations

In the solution of many problems it is required to find the functions y1 ( x) , y 2 ( x) …etc. of the same argument x where the functions y1 , y 2 …satisfy a system of differential equations as in the following: dy1 = f1 ( x, y1 , y 2 ) ………………….(1) dx dy 2 = f 2 ( x, y1 , y 2 ) …………………..(2) dx

Here, we have two coupled differential equations. The task is to find out the solutions y1 ( x) and y 2 ( x) . Example #1

Integrate the system: dy = y + z + x ……………….(1) dx dz = −4 y − 3z + 2 x ………….(2) dx with the initial conditions y ( x = 0) = 1 , z ( x = 0) = 0 . To find solutions: Differentiate equation (1) d 2 y dy dz = + + 1 …………...(3) dx 2 dx dx Put back equation (2) into equation (3): d 2 y dy = + (−4 y − 3z + 2 x) + 1 dx 2 dx

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

dy  dy  [From eqn. (1)] − 4 y + 2 x + 1 − 3 − y − x  dx  dx  d2y dy + 2 + y = 5 x + 1 …………………….(4). 2 dx dx

= ⇒

Therefore, we have got a 2nd order differential equation with constant coefficients in above. The general solution of equation (4) is y = (c1 + c 2 x)e − x + 5 x − 9 . dy − y−x dx = (c 2 − 2c1 − 2c 2 x)e − x − 6 x + 14 . We use the initial conditions to find out the arbitrary constants c1 and c 2 . We find c1 = 10 and c 2 = 6 . Thus we get the following solutions which satisfy the given set of equations and the initial conditions: y = (10 + 6 x)e − x + 5 x − 9 z = (−14 − 12 x)e − x − 6 x + 14 .

Now we have,

z=

Example #2

Integrate the system:

To find solutions:

dx = y + z ……………….(1) dt dy = x + z ……………….(2) dt dz = x + y ……………….(3) dt

Differentiating the first equation with respect to t , we find d 2 x dy dz = + = ( x + z) + ( x + y) = 2 x + y + z dt dt dt 2 Eliminating the variables y and z , we get d 2 x dx − − 2x = 0 . dt 2 dt So we get a 2nd order homogeneous equation, the general solution of which is x = c1e −t + c 2 e 2t . From the above we find

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dx = −c1e −t + 2c 2 e 2t dt dx ∴ y= − z = − c1e −t + 2c 2 e 2t − z dt Now we put the above relation into equation (3) and get dz = x+ y dt = (c1e −t + c 2 e 2t ) + (− c1e −t + 2c 2 e 2t ) − z dz ∴ + z = 3c 2 e 2t dt The above is a first order differential equation of z . The solution is z = c3 e − t + c 2 e 2t . Therefore, y = −c1e − t + 2c 2 e 2t − c3 e −t + c 2 e 2t

= − (c1 + c3 )e −t + c 2 e 2t . Lecture-10 Physical Interpretation:

Let us consider the vibration in a Mechanical System. Example: Oscillation of a spring y

M 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 dy opposite to that of the motion. The force is proportional to velocity: − λ . dt [Resistance force or Damping force]

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We can write the differential equation of the motion of the load on the spring. By Newton’s 2nd law we have d2y dy M 2 = −ky − λ . dt dt Thus we have the following homogeneous 2nd order differential equation of the familiar form: y // + py / + qy = 0 ………………….(1) k λ Where p = , q= . M M If the spring-load system is further acted on by an external force f (t ) then the corresponding differential equation can be written as y // + py / + qy = f (t ) ………………..(2) Equation (1) is called an equation of FREE OSCILLATIONS and equation (2) is called an equation of FORCED OSCILLATIONS. FREE OSCILLATIONS y // + py / + qy = 0 Corresponding auxiliary equation is k 2 + pk + q = 0 . Two roots are: k1 = −

p + 2

p2 −q 4

k2 = −

p − 2

p2 −q 4

CASE-I

p2 > q (Case of over damping) 4 Then the roots k1 and k 2 are negative numbers. The general solution: y = c1e k1t + c2e k 2 t ( k1 < 0 , k2 < 0 ) 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

p2 = q Then the roots are equal. 4 The general solution is y = (c1 + c 2 t )e − p / 2.t .

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Here the displacement ( y ) approaches zero as t → ∞ (long time limit) but comparatively slower than that of case-II. CASE-III p = 0 (No resistance, No damping) Equation (1) takes the form y // + qy = 0 . y = c1 cos ωt + c 2 sin ωt , where ω = q . The general solution is: If we write c1 = A sin φ 0 and c 2 = A cos φ 0 , the solution becomes y = A sin(ωt + φ 0 ) . This is a solution of Simple Harmonic motion.

CASE-IV (Damped Oscillations)

p2 < 0, β = q − p 2 / 4 . 2 The general solution has the form: y = e αt (c1 cos βt + c 2 sin βt ) Or y = Ae αt sin( β t + φ 0 ) .

p ≠ 0,

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 e αt term.

The envelope of the peaks falls off exponentially.

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

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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: y ′′ + p ( x) y ′ + q( x) y = f ( x) .

Assume the following form of the solution: y = vy1 + v 2 y 2 , (1) where y1 and y 2 are two independent solutions of the corresponding homogeneous equations. The coefficients v1 and v 2 are two unknown functions. The function f ( x) is considered to be continuous in some range. Therefore, we can write y1′′ + p ( x) y1′ + q ( x) y1 = 0 ………………….(2) y 2′′ + p ( x) y 2′ + q ( x) y 2 = 0 ...……………..…(3)

Now from (1), we can have y ′ = v1′ y1 + v1 y1′ + v ′2 y 2 + v 2 y 2′ = (v1 y1′ + v 2 y 2′ ) + (v1′ y 1 +v ′2 y 2 ) .

(4)

To simplify the calculation of y ′′ , we choose v1 and v 2 so that v1′ y1 + v 2′ y 2 = 0 . (5) The final solution corresponding to the given equation will be subject to the condition we choose. The equation (4) now becomes y ′ = v1 y1′ + v 2 y 2′ (6) Now differentiating equation (5) once again with respect to x , we obtain y ′′ = v1′ y1′ + v1 y1′′ + v ′2 y ′2 + v 2 y 2′′ .

(7)

Substituting equations (1), (6) and (7) into the original equation, we obtain v1′ y1′ + v1 y1′′ + v 2′ y ′2 + v 2 y 2′′ + p ( x)v1 y1′ + p ( x)v 2 y 2′ + q ( x)v1 y1 + q ( x)v 2 y 2 = f ( x)

⇒ v1 ( y1′′ + p ( x) y1′ + q ( x) y1 ) + v 2 ( y 2′′ + p ( x) y 2′ + q ( x) y 2 ) + v1′ y1′ + v 2′ y 2′ = f ( x) .

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The two expressions in parentheses in the last equation vanish as they are essentially the solutions of the homogeneous equation. Hence we now have v1′ y1′ + v 2′ y ′2 = f ( x) . (8) Solving equations (5) and (8) for v1′ and v 2′ , we obtain 0 f ( x)

v1′ =

y2 y ′2

∆

=−

y 2 f ( x) ∆

and v 2′ =

0 f ( x)

y1 y1′

∆

=

y1 f ( x) , ∆

where ∆=

y1 y1′

y2 = ( y1 y ′2 − y1′ y 2 ) . y 2′

The determinant ∆ is called the Wronskian W ( y1 , y 2 ) of y1 and y 2 , The quantities v1 and v 2 can be found out from the above by integration. Therefore, we obtain the following solution: y = − y1 ∫

f ( x) y 2 f ( x) y1 dx + y 2 ∫ dx . W ( y1 , y 2 ) W ( y1 , y 2 )

The above is the general solution of the given equation. This method is useful when the two independent solutions y1 and y 2 can not be obtained easily. Example #1

To solve: y ′′ + y = sec x . [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 y1 = sin x and y 2 = cos x .

∴ The Wronskian is W ( y1 , y 2 ) = ∆ =

y1 y1′

y2 sin x cos x = = −1 . y 2′ cos x − sin x

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Therefore, the particular integral is y p = sin x ∫ sec x. cos xdx − cos x ∫ sec x. sin xdx sin x dx cos x = − x sin x + cos x log(cos x) .

= − sin x ∫ dx − cos x ∫

The complete solution is thus y = A sin x + B cos x − x sin x + cos x log(cos x). Example #2

To solve: x 2 y ′′ − 2 xy ′ + 2 y = x ln x ; (x ≠ 0 ) Soln: The standard form for the equation is

y ′′ − where p ( x) = −

2 2 ln x y′ + 2 y = , x x x

2 x

2 x2 ln x f ( x) = x Let us take two independent solutions of the corresponding homogeneous equation by investigation * . We have y1 = x and y 2 = x 2 . Hence we have y y2 x x2 = = x2 . W ( y1 , y 2 ) = ∆ = 1 y1′ y 2′ 1 2x Thus the general solution is  ln x   ln x  x2  x   x  x    2 dx y = − x∫ dx + x ∫ x2 x2 ln x ln x = − x∫ dx + x 2 ∫ 2 dx x x 1   ln x 1  2 = − x  (ln x ) + c1  + x 2 − − + c2  x 2   x  1  2 = − x  (ln x ) + ln x + 1 − c1 x + c 2 x 2 . 2  q ( x) =

*

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.

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Cauchy-Euler Equation:

If we have an equation of the following type x 2 y ′′ + axy ′ + by = 0 , we search for the solutions of the algebraic form y ( x) = x m . In the above example, we have the homogeneous equation : x 2 y ′′ − 2 xy ′ + 2 y = 0 To solve, we look for the solution of the form, y ( x) = x m . Thus we shall have y ′ = mx m −1 and y ′′ = m(m − 1) x m − 2 . Hence the corresponding auxiliary equation becomes m 2 − 3m + 2 = 0 . The two roots are m1 = 1 and m2 = 2 . Therefore, we have two independent solutions y1 = x and y 2 = x 2 . -----------------------------