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Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 12 textbook solution.

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Answer : \text { (a) } y=x e^{x+c}

Hint: Use variable separable method i.e. take the y terms in one side and x terms in the other.

Given : \frac{d y}{d x}-\frac{y(x+1)}{x}=0

Explanation : \frac{d y}{d x}=\frac{y(x+1)}{x}

\begin{aligned} &\frac{d y}{y}=d x\left(\frac{x+1}{x}\right) \\ &\frac{d y}{y}=\left(1+\frac{1}{x}\right) d x \end{aligned}

Integrate both sides

\begin{aligned} &\log y=x+\log x+C \\ &\log y-\log x=x+C \\ &\log \frac{y}{x}=x+C \end{aligned}

Take exponential both sides we get

\begin{aligned} &\frac{y}{x}=e^{x+c} \\ &y=x e^{x+C} \end{aligned}

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