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# Second Order Differential Equation

Second order differential equation involves the process of differentiating the given algebraic function twice with respect to the given variable. Generally, the second order differential is discussed in calculus. It mainly helps to find the rate of change of the given function with respect to the change in the input.

 Related Calculators Second Order Differential Equation Calculator Differential Equation Calculator System of Differential Equations Solver Order from least to Greatest

## Second Order Linear Differential Equation

We know that the differential equation is linear, if all the derivatives which is present in it are linear. So, we can say that the differential equation is in the following form:

$A(x)$$\frac{d^{2}y}{dx^{2}}$$ + B(x) $$\frac{dy}{dx}$$ + C(x) y = G(x)$

Where A, B, C and G are continuous functions of x. If G(x) = 0, then the differential equation is called as homogeneous linear differential equation.

A second order differential equation which does not fulfill the condition of linear equation is called the second order non-linear differential equation. To find the solution of non-linear differential equation, we have few methods those depends on the equation having particular symmetric.

## Solving Second Order Differential Equations

Let us have a second order differential equation of the form of $A(x)$$\frac{d^{2}y}{dx^{2}}$$ + B(x) $$\frac{dy}{dx}$$ + C(x) y = 0$

For finding the solution, we have replace $\frac{dy}{dx}$ by $m$, $\frac{d^{2}y}{dx^{2}}$ by $m^2$ etc. to make it in the auxiliary equation. Then, the solution is called complimentary function of the differential equation.

Case 1: b2 - 4ac > 0:

If we have the differential equation in the form of Ay" + B y' + C y = 0, and after replacing $\frac{dy}{dx}$ by $m$, $\frac{d^{2}y}{dx^{2}}$ by $m^2$, we get Am2 + Bm + C = 0. And, if the roots of the equation are m1 and m2 i.e. if the roots of auxiliary equation are real and different, then the solution of the given differential equation is $y = c_{1}e^{m_{1}x} + c_{2}e^{m_{2}x}$

Case 2: b2 - 4ac = 0:

If we have the differential equation in the form of Ay" + B y' + C y = 0, and after replacing $\frac{dy}{dx}$ by $m$, $\frac{d^{2}y}{dx^{2}}$ by $m^2$, we get Am2 + Bm + C = 0. And, if the roots of the equation are m1 and m1 i.e. if the roots of auxiliary equation are real and same(equal), then the solution of the given differential equation is $y = c_{1}e^{m_{1}x} + x c_{2}e^{m_{1}x}= \left ( c_{1}+ c_{2}x \right )e^{m_{1}x}$

Case 3: b2- 4ac < 0:

If we have the differential equation in the form of Ay" + B y' + C y = 0, and after replacing $\frac{dy}{dx}$ by $m$, $\frac{d^{2}y}{dx^{2}}$ by $m^2$, we get Am2 + Bm + C = 0. And, if the roots of the equation are complex numbers and real i.e. $m_{1} = \alpha + i\beta , m_{2} = \alpha - i\beta$, then the solution of the given differential equation is $y = c_{1}e^{\alpha + i\beta}x+ c_{2}e^{\alpha - i\beta}x$

= $c_{1}e^{\alpha x}\left ( cos \beta x + sin \beta x \right )$

= $e^{\alpha x} \left ( c_{1}cos \beta x + c_{2}sin \beta x \right )$

Let us have a second order differential equation in the form of $A(x)$$\frac{d^{2}y}{dx^{2}}$$ + B(x) $$\frac{dy}{dx}$$ + C(x) y = Q(x)$

Then, for solving this type of differential equation, we have to find complimentary function as well as the particular integral. For the particular integral, we have

Case 1:

If we have Q(x) = emx and say f(D) = AD2 + BD + C = 0,then particular integral is given as follows:

P.I. = $\frac{1}{f(D)}$ $e^{mx}$ = $\frac{1}{f(m)}$ $e^{mx}$ where $f(m) \neq 0$.

1. If $f(m) = 0$ and $f'(m) \neq 0$, then P.I. is $\frac{x e^{mx}}{f'(m)}$
2. If $f(m) = f'(m) = 0$ and $f"(m) \neq 0$, then P.I. is $\frac{x e^{mx}}{f"(m)}$
Case 2:

If we have $Q(x) = \sin (mx)$ or $\cos(mx)$, then we can express $f(D)$ in terms of $D^2$ and after that replace $D^2$ by $-m^2$
P.I. is $\frac{1}{f(D)}$ $\sin (mx)$

= $\frac{1}{\phi(D^{2})}$ $\sin (m)$

= $\frac{1}{\phi(-m^{2})}$ $\sin (m)$ where $\phi(-m^2) \neq =0$.

Sometimes, we don't have $D^2$ in $f(D)$, then we will try to get $D$ and $D^2$ in $f(D)$.
Let $f(D) = D^2 + 2D + 1$ and $Q(x) = \sin 3x$, then

P.I. is $\frac{1}{D^{2}+ 2D + 1}$$\sin 3x Replace D^2 by (-32) = \frac{1}{-9 + 2 D + 1}$$ \sin 3x$

= $\frac{1}{2D - 8}$$\sin 3x To get the term D^2, multiply and divide by conjugate of (2D - 8) i.e. by (2D + 8), we get = \frac{2D + 8}{(2D - 8)(2D + 8)}$$\sin 3x$

= $\frac{2D + 8}{4D^{2} - 64}$$\sin 3x = \frac{2D + 8}{4 (-9) - 64}$$\sin 3x$

= $\frac{-1}{100}$$[2D(\sin 3x) + 8 \sin 3x] = \frac{-1}{100}$$[6 \cos 3x + 8 \sin 3x]$

## Second Order Differential Equation Examples

Given below are some of the examples on second order differential equations.

### Solved Examples

Question 1: Calculate (D2 - 7D + 10) y = 0
Solution:
We have (D2 - 7D + 10) y = 0
Replace D by m to find the root of the auxiliary equation, then
m2 - 7 m + 10 = 0
(m - 5) (m - 2) = 0
m = 2, 5
So, the solution of the given equation is y = c1e2x + c2 e5x

Question 2: Solve (D2 - 4D + 4) y = e3x
Solution:
In this equation, we have Q(x) = e3x
Hence, the solution of the equation is y = C.F. + P.I.
Now, from auxiliary equation, we get roots of the equation.
Hence, m2 - 4m + 4 = 0
Therefore, m = 2,2.
C.F. = (c1 + xc2)e2x

For P.I. = $\frac{1}{D^{2} - 4D + 4}$$e^{2x} Here, f(4) = f'(4) = 0 Then, P.I. = \frac{x^{2}e^{2x}}{2} Therefore, the complete solution of the equation is y = (c_1 + xc_2)e^{2x} + \frac{x^{2}e^{2x}}{2} Question 3: Solve (D2 - 64)y = cos 3x Solution: Given that (D2 - 64) y = cos 3x Therefore, complete solution of the equation is y = C.F. + P.I. The auxiliary equation is m2 - 64 = 0 (m + 8)(m - 8) = 0 m = 8, -8 C.F.= c1 e8x + c2 e-8x Now, P.I. = \frac{1}{f(D)}$$\cos 3x$

= $\frac{1}{D^{2}- 64}$$\cos 3x Replacing D2 by -32, we get = \frac{1}{-9 - 64}$$\cos 3x$

= $\frac{-1}{73}$$\cos 3x Hence, the complete solution of the equation is y = c_1 e^{8x} + c_2 e^{-8x} +$$ \frac{-1}{73}$$\cos 3x Question 4: Solve (D2 + 4D + 4)y = cos 2x Solution: We have (D2 + 4D + 4)y = cos 2x Therefore complete solution of the equation is y = C.F. + P.I. The auxiliary equation is m2 + 4m + 4 = 0 (m - 2)2 = 0 m = -2, -2 C.F.= (c1 + x c2)e-2x Now, P.I. = \frac{1}{f(D)}$$\cos 2x$

= $\frac{1}{D^{2} + 4D + 4}$$\cos 2x Replace D2 by -22, then = \frac{1}{-4 + 4D + 4}$$\cos 2x$

= $\frac{1}{4D}$$\cos 2x Here, \frac{1}{D} indicates the sign of integration, hence we get = \frac{1}{4} [\frac{\sin 2x}{2}] = \frac{\sin 2x}{8} So, the complete solution of the equation is y = (c_1 + x c_2)e^{-2x} +$$\frac{\sin 2x}{8}$

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