Second Order Linear Differential Equation

When we calculate rate of any function with respect to any variable, then this procedure is called as a differentiation. For example y = f(x), then if we differentiate y with respect to x, it produces dy/dx, which is called as a derivative of y with respect to x, and it produces rate of f(x) with respect to x. An equation which contains dy/dx is called as a first order derivative equation because here we first time differentiate y with respect to x and when we differentiate dy/dx it produces second order derivative (d²y/dx²). So, when we differentiate any function with respect to particular variable, it produces second order derivative and a linear equation which contains second order derivative is called as a second order linear differential equation. Now we discuss some second order linear differential equation example–

Like if y = x + tan x, then second order linear differential equation is cos² x (d²y/dx²) – 2y + 2x = 0.

For proving this kind of second order linear differential equation, we use following steps –

Step 1: First of all we assume function as a variable like y = f(x).

Step 2: After assuming the variable, we differentiate that variable with respect to any particular variable like we differentiate y with respect to x, it produces dy/dx.

Step 3: According to the equation we again differentiate variable with respect to that particular variable like if we differentiate dy/dx with respect to x again, it produces (d²y/dx²).

Step 4: After all differentiate process; we put the values of dy/dx and d²y/dx² in that equation, which we want to prove.

For understand these steps we take some examples -

Example 1: If y = sin^-1 x, then prove (d²y/dx²) = x/(1 – x²)^3/2

Solution:

Step 1: In above question, given that y = sin ^-1 x

 Step 2: On differentiation of y with respect to x, we get

                   dy/dx = 1/√(1 – x²)                           ……………..equation(1)

  Step 3:  On differentiation again with respect to x, we get:

                   d²y/dx² = d/dx(1/√(1 – x²)) = d/dx[(1 – x²)^-1/2],

=>  d²y/dx² = -1/2 * (1 – x²)^-3/2 * d/dx(1 – x²),

=>   d²y/dx² = -1/2 * (1 – x²)^-3/2 * (-2x),

=>   d²y/dx² = x/(1 – x²)^3/2                                 Hence proved.        

Example 2: If y = A cos nx + B sin nx, then prove (d²y/dx²) + n²y = 0

Solution:

Step 1: In above question, given that y = A cos nx + B sin nx

Step 2: On differentiation of y with respect to x, we get

                dy/dx = - An sin nx + Bn cos nx                           ……………..equation(1)

Step 3:  On differentiation again with respect to x, we get

                   d²y/dx² = d/dx(- An sin nx + Bn cos nx ) 

=>  d²y/dx² = - An² cos nx + Bn² sin nx

=>  d²y/dx² = - n² (A cos nx + B sin nx)

=>  ( d²y/dx²) = - n²y

=>   (d²y/dx²) + n²y = 0                                                 Hence proved.

Example 3: If y = A e^mx + B e^nx , then prove d²y/dx² – (m + n)dy/dx + mny = 0

Solution:

Step 1: In above question, given that y = A e^mx + B e^nx

Step 2: On differentiation of y with respect to x, we get

                dy/dx = Am e^mx + Bn e^nx                           ……………..equation(1)

Step 3:  On differentiation again with respect to x, we get

                   d²y/dx² = d/dx(Am e^mx + Bn e^nx ) =

=>  d²y/dx² =  Am² e^mx + Bn² e^nx             ………………equation(2)

Step 4: After all differentiate part we manage our equation

                  d²y/dx² – (m + n)dy/dx + mny = 0

                 we put d²y/dx² from eq(2) and dy/dx from eq(1) in left hand part of above equation    

                 Am² e^mx + Bn² e^nx  - (m + n)* (Am e^mx + Bn e^nx) + mn * (e^mx + B e^nx)                                     

=>  Am² e^mx + Bn² e^nx  -  Am² e^mx – Bn² e^nx  - mn * (e^mx + B e^nx) + mn * (e^mx + B e^nx)                                    

=>    0 =  Rhs

So, d²y/dx² – (m + n)dy/dx + mny = 0                                               Hence proved.

Example 4: If y = A cos ( log x) + B sin ( log x), then prove x² * (d²y/dx²) +  x dy/dx + y = 0

Solution:

Step 1: In above question, given that y = A cos ( log x) + B sin ( log x)

Step 2: On differentiation of y with respect to x, we get

                dy/dx = -1/x * A sin (log x) + 1/x * B cos(log x)   

               =>  x *(dy/dx) = - A sin (log x) + B cos(log x)         ……………..equation(1)

Step 3:  On differentiation of eq(1) with respect to x, we get

                   x * (d²y/dx²) = -A cos (log x)/x – B sin(log x)/x

=>  x² * (d²y/dx²) + x dy/dx=  -(A cos (log x) + B sin (log x))

=>  x² * (d²y/dx²) + x dy/dx = -y

=>  x² * (d²y/dx²) +  x dy/dx + y = 0             Hence proved.

Example 5: If y = tan x + sec x, then prove (d²y/dx²) = cos x/(1 – sin x)²

Solution:

Step 1: In above question, given that y = tan x + sec x

Step 2: On differentiation of y with respect to x, we get

                dy/dx = sec² x + sec x. tan x

         =>  dy/dx = 1/cos² x + sin x/ cos² x

         =>  dy/dx = (1 + sin x)/ cos² x

         =>  dy/dx = (1 + sin x)/ (1 – sin² x)

         =>  dy/dx = 1/(1-sin x)             ……………..equation(1)

Step 3:  On differentiation of eq(1) with respect to x, we get

                (d²y/dx²) = d/dx(1/(1 – sin x)) =  d/dx((1 – sin x)^-1 )

=>  d²y/dx² = (-1)(1- sin x)^-2 d/dx(1 – sin x)

=>   d²y/dx² = (-1)(1- sin x)^-2 (- cos x)

=>  (d²y/dx²) = cos x/(1 – sin x)            Hence proved.

These are some examples which show how we calculate linear second order differential equation and how we can prove linear second order differential equation.