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Explain solution for  RD Sharma Class 12 Chapter 21 Differential Equation Exercise Multiple choice Question Question 35 maths textbook solution.

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Answer :  \text { (b) } \tan ^{-1} \frac{y}{x}=\log x+C

Hint: Let v=\frac{y}{x}

Given : \frac{d y}{d x}=\frac{x^{2}+x y+y^{2}}{x^{2}}

Explanation :  \frac{d y}{d x}=\frac{x^{2}}{x^{2}}+\frac{x y}{x^{2}}+\frac{y^{2}}{x^{2}}

\Rightarrow \frac{d y}{d x}=1+\frac{y}{x}+\frac{y^{2}}{x^{2}}                         ...(i)

Let y=vx

\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}

Put in (i) we get

\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=1+v+v^{2} \\ &\Rightarrow x \frac{d v}{d x}=1+v^{2} \\ &\Rightarrow \frac{d v}{1+v^{2}}=\frac{d x}{x} \end{aligned}

Integrate on both sides

\begin{aligned} &\Rightarrow \tan ^{-1} v=\log x+C \\ &\Rightarrow \tan ^{-1} \frac{y}{x}=\log x+C \quad\left[v=\frac{y}{x}\right] \end{aligned}

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