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Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 45 sub question (viii) maths

Answers (1)


Answer: \tan ^{-1} y=\frac{x^{2}}{2}+x

Hint: Separate the terms of x and y and then integrate them.

Given: \frac{d y}{d x}=1+x+y^{2}+x y^{2}, \text { when } y=0, x=0


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

        Integrating both sides

        \begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int(1+x) d x \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{2}}{2}+c \end{aligned}        ..............(1)

        Given that when y=0, x=0

        \therefore \tan ^{-1}(0)=0+\frac{0}{2}+c \Rightarrow c=0

        Put in (1) we get

        \begin{aligned} &\tan ^{-1} y=x+\frac{x^{2}}{2}+0 \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{2}}{2} \end{aligned}

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