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

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

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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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