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Variation of Parameters

Probability Introduction :

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

1) http://www.mathmaa.com/support-files/variationofparameterssample.pdf


2)http://www.mathmaa.com/support-files/sampleprob2variationofparameters.pdf

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 \[\large y=c_{1}e^{-x}+c_{2}xe^{-3x}-\ln \left | x^{2}+1 \right |e^{-3x}+2\tan^{-1}(x)e^{-3x}\]




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