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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 37 subquestion (iv)

Answers (1)

Answer:  \begin{aligned} & y=-e^{-x}+\frac{x}{e} \\ & \end{aligned}

Give:  x \frac{d y}{d x}-y=(x+1) e^{-2 x}, y(1)=0

Hint: Using integration by parts and  \int \frac{1}{x}dx

Explanation:  x \frac{d y}{d x}-y=(x+1) e^{-x}

Divide by x , we get

 \begin{aligned} &=\frac{d y}{d x}-y\left(\frac{1}{x}\right)=\frac{(x+1)}{x} e^{-x} \\ &=\frac{d y}{d x}+y\left(\frac{-1}{x}\right)=\frac{(x+1)}{x} e^{-x} \end{aligned}

This is a linear differential equation of the form

 \begin{aligned} &\frac{d y}{d x}+P x=Q \\ &P=-\frac{1}{x} \text { and } Q=\frac{(x+1)}{x} e^{-x} \end{aligned}

The integrating factor   of this differential equation is

 \begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{1}{x} d x} \\ &=e^{-\int \frac{1}{x} d x} \\ &=e^{-\log |x|} \quad\left[\int \frac{1}{x} d x=\log x+C\right] \end{aligned}

Hence, the solution is

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y\left(\frac{1}{x}\right)=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \\ &=\frac{y}{x}=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \end{aligned}

\begin{aligned} &=\frac{y}{x}=I_{\prime}+C \ldots(i) \\ &=I_{1}=\int \frac{(x+1)}{x} e^{-x} \frac{1}{x} d x+C \\ &=\frac{1}{x} \frac{e^{-x}}{x}-\int\left(\frac{-1}{x^{2}}\right) \frac{e^{-x}}{-1} d x+\int \frac{e^{-x}}{x^{2}} d x \\ &=-\frac{e^{-x}}{x}-\int \frac{e^{-x}}{x^{2}} d x+\int \frac{e^{-x}}{x^{2}} d x \\ &=-\frac{e^{-x}}{x} \end{aligned}

By  \frac{y}{x}=-\frac{e^{-x}}{x}+C

Multiply by x

=y=-e^{-x}+C x \ldots(i i) 

Now  y(1)=0 \text { when } x=1, y=0

\begin{aligned} &=0=-e^{-1}+C(1) \\ &=0=-\frac{1}{e}+C \\ &=C=\frac{1}{e} \end{aligned} 


Put in (ii)

 \begin{aligned} &=y=-e^{-x}+\frac{1}{e} x \\ &=y=-e^{-}+\frac{x}{e} \end{aligned}

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