online user counter

MATHMAA

# Variation of Parameters

Concept of Variation of Parameters:

General solution of $\large a_{2}\frac{d^{2}y}{dx^{2}}+a_{1}\frac{\mathrm{d} y}{\mathrm{d} x}+a_{0}y=f(x)$by the method of variation of parameters.

Given linear differential equation is $\large a_{2}\frac{d^{2}y}{dx^{2}}+a_{1}\frac{\mathrm{d} y}{\mathrm{d} x}+a_{0}y=f(x)\cdots (1)$ where $\large a_{2}\neq 0$ a1 , a0 are function of x or real constants and f(x) is only a function of x. The homogeneous equations corresponding to (1) is $\large a_{2}\frac{d^{2}y}{dx^{2}}+a_{1}\frac{\mathrm{d} y}{\mathrm{d} x}+a_{0}y=0\cdots (2)$

Let $\large y=c_{1}y_{1}+c_{2}y_{2}$ c2y2 be the general solution of (2) where y1 and y2 are two L.I. solutions of (2) and c1 ,c2 are real constants.

Let particular solutions of (1) be yp = u1y1+u2y2 ------(3) which is obtained from yc of (1) by replacing c1 ,c2 by u1 ,u2 which are also some functions of x and whose values are to be determined.

$\large {u_{1}}'=\frac{W_{1}}{W}=\frac{-y_{2}f(x)}{W}$and $\large {u_{2}}'=\frac{W_{2}}{W}=\frac{y_{1}f(x)}{W}\cdots(4)$ where $\large W=\begin{vmatrix} y_{1} &y_{2} \\ {y_{1}}'& {y_{2}}' \end{vmatrix}$$\large W_{1}=\begin{vmatrix} 0 &y_{2} \\ f(x) & {y_{2}}' \end{vmatrix}$$\large W_{2}=\begin{vmatrix} y_{1} &0 \\ {y_{1}}' & f(x) \end{vmatrix}$

Integrate u1' and u2' to find u1 and u2.
Therefore Solution of the given equation is $\large y=y_{c}+y_{p} .$

Concept of Variation of Parameters:

Click on the link given BELOW

http://www.mathmaa.com/support-files/conceptofvariationofparameters.pdf

Sample problem of Variation of Parameters:

Click on the link given BELOW

Q1) Solve differential equations by using Variations of parameters.                     ${y}''+6{y}'+9y=\frac{2e^{-3x}}{x^{2}+1}$ .

Sol:

${y}''+6{y}'+9y=\frac{2e^{-3x}}{x^{2}+1}$

A.E.

$m^{2} + 6m + 9 =0$

$\left (m+3 \right )^{2}= 0$

m= -3, -3

$y_{c}=c_{1}e^{-3x}+c_{2}xe^{-3x}$

..........(1)

$y_{1}=e^{-3x}$

and

$y_{2}=xe^{-3x}$

$y_{p}=u_{1}y_{1}+u_{2}y_{2}$

................(2)

$W(y_{1},y_{2})=\begin{vmatrix} e^{-3x} &xe^{-3x} \\ -3e^{-3x}& (1-3x)e^{-3x} \end{vmatrix}$

$= e^{-6x}\neq 0$

$W_{1}=\begin{vmatrix} 0 & xe^{-3x}\\ \frac{2e^{-3x}}{x^{2}+1} & (1-3x)e^{-3x} \end{vmatrix}$

$=\frac{-2xe^{-6x}}{x^{2}+1}$
$W_{2}=\begin{vmatrix} e^{-3x} & 0\\ -3e^{-3x} & \frac{2e^{-3x}}{x^{2}+1} \end{vmatrix}=\frac{2e^{-6x}}{x^{2}+1}$

$\large {u_{1}}'=\frac{W_{1}}{W}=\frac{-2x}{x^2+1}$
$\large u_{1}=-\ln \left | x^{2}+1 \right |$
$\large {u_{2}}'=\frac{W_{2}}{W}=\frac{2}{x^{2}+1}$
$\large u_{2}=2\tan^{-1}(x)$
$\large y_{p}=u_{1}y_{1}+u_{2}y_{2}$
$\large y_{p}=-\ln \left | x^{2}+1 \right |e^{-3x}+2\tan^{-1}(x)e^{-3x}.....(3)$
$\large y=y_{c}+y_{p}$
Therefore General Solution is $\large y=c_{1}e^{-x}+c_{2}xe^{-3x}-\ln \left | x^{2}+1 \right |e^{-3x}+2\tan^{-1}(x)e^{-3x}$