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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 5

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Answer: $\frac{y}{x}=\log |x|+C$

Hint: To solve this equation we use

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

Solution: $\frac{d y}{d x}=1+\frac{y}{x}$

$\begin{aligned} &\frac{d y}{d x}-\frac{1}{x} y=1 \\ \end{aligned}$

$\frac{d y}{d x}-\frac{1}{x} y=1 \\$

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

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

$\text { If }=e^{\int P d x} \\$

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

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

$=x^{-1} \\$

$=\frac{1}{x}$

$\begin{aligned} &y \times \frac{1}{x}=\int 1 \times \frac{1}{x} d x+C \\ & \end{aligned}$

$\frac{y}{x}=\log x+C \\$

$\frac{y}{x}=\log |x|+C$

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