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  • Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (iv) maths textbook solution.

Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 37 Subquestion (iv) maths textbook solution.

Answers (1)

Answer : y=-e^{-x}+\frac{x}{e}

Given : 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} d x

Explanation : x \frac{d y}{d x}-y=(x+1) e^{-2 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 If  of this differential equation is

          \begin{aligned} &\text { If }=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, is the solution is

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

            \begin{aligned} &=-\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 (i) \frac{y}{x}=-\frac{e^{-x}}{x}+C

 Multiply by x

               \begin{aligned} &=y=-e^{-x}+C x \ldots(i i) \\ &\text{Now}\; y(1)=0 \text { when } x=1, y=0 \\ &=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} 

Posted by

infoexpert23

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