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# 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 emx + B enx , then prove d2y/dx2 – (m + n)dy/dx + mny = 0

Solution:

Step 1: In above question, given that y = A emx + B enx

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

dy/dx = Am emx + Bn enx                           ……………..equation(1)

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

d2y/dx2 = d/dx(Am emx + Bn enx ) =

=>  d2y/dx2 =  Am2 emx + Bn2 enx             ………………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 emx + Bn2 enx  - (m + n)* (Am emx + Bn enx) + mn * (emx + B enx)

=>  Am2 emx + Bn2 enx  -  Am2 emx – Bn2 enx  - mn * (emx + B enx) + mn * (emx + B enx)

=>    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.

## Solve Second Order Differential Equations

What is differentiation? In differentiation Calculus, differentiation is a process of finding derivative of function where derivative means rate of change with respect to some variable like derivative of f(x) with respect to x is:

d   (f(x).

dx

## Application of Linear second order differential equations

The most general form of the linear second order differential equation with constant coefficients is:                           P (dy2/dt2 )+ Q ( dy/dt) + R (y) = f(t)                                                        (1)

Where P, Q, R are the constants.

The other form of linear second order differential equati...Read More

• ## Second Derivative Calculator

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