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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 9

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Answer: $4 x y=2 x^{2} \log |x|-x^{2}+C$

Hint: To solve this equation we will use differentiation method.

Give: $\frac{d y}{d x}+\frac{y}{x}=\log x$

Solution: $\frac{d y}{d x}+\frac{y}{x}=\log x$

$\frac{d y}{d x}+P y=Q$

$\begin{aligned} &P=\frac{1}{x^{\prime}} Q=\log x \\ & \end{aligned}$

$I\! f=e^{\int \frac{1}{x} d x} \\$

$=e^{\log x} \\$

$=x \\$

$y I\! f=\int Q I\! f d x \\$

$y x=\int x \log x d x+C$

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

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

$=\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C \\$

$=y=\frac{x \log x}{2}-\frac{x}{4}+\frac{C}{x} \\$

$=4 x y=2 x^{2} \log x-x+C$

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