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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 10

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Answer: $y=e^{x}+C x$

Hint: To solve this equation we will use differentiate separately.

Give: $\begin{aligned} &x \frac{d y}{d x}-y=(x-1) e^{x} \\ & \end{aligned}$

Solution: $\frac{d y}{d x}-\frac{1}{x} y \frac{(x-1) e^{x}}{x}$

$\begin{aligned} &\quad \frac{d y}{d x}+P y=Q \\ & \end{aligned}$

$P=\frac{-1}{x}, Q=\frac{(x-1) e^{x}}{x} \\$

$I \! f=e^{\int P d x} \\$

$=e^{-\int \frac{1}{x} d x}$

$\begin{aligned} &=e^{\log x^{-1}} \\ & \end{aligned}$

$=x^{-1} \\$

$y I \! f=\int Q I\! f+C \\$

$=y \frac{1}{x}=\int(x-1) e^{x} \frac{1}{x} d x+C \\$

$=\frac{y}{x}=\int \frac{x e^{x}}{x^{2}} d x-\int \frac{e^{x}}{x^{2}} d x+C$

$\begin{aligned} &=\frac{1}{x} e^{x}-\int \frac{1}{x^{2}} e^{x} d x-\int \frac{1}{x^{2}} e^{x} d x \\ \end{aligned}$

$=\frac{1}{x} e^{x} \\$

$=\frac{y}{x}=\frac{e^{x}}{x}+C \\$

$=y=e^{x}+C x$

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