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Need solution for  RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 16 textbook solution.

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Answer : (d) y=\tan \left(x+\frac{x^{2}}{2}\right)

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

Given :  \frac{d y}{d x}=1+x+y^{2}+x y^{2}, \quad y(0)=0

Explanation : \frac{d y}{d x}=1+x+y^{2}+x y^{2}

\begin{aligned} &\frac{d y}{d x}=1+x+y^{2}(1+x) \\ &\frac{d y}{d x}=(1+x)\left(1+y^{2}\right) \\ &\frac{d y}{1+y^{2}}=(1+x) d x \end{aligned}

\begin{aligned} &\tan ^{-1} y=x+\frac{x^{2}}{2}+C\\ &\text { Also we have } y(0)=0\\ &\tan ^{-1}(0)=0+\frac{0}{2}+C\\ &C=0 \end{aligned}

\begin{aligned} &\text { So, } \tan ^{-1} y=x+\frac{x^{2}}{2} \\ &y=\tan \left(x+\frac{x^{2}}{2}\right) \end{aligned}

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