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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 45 sub question (vi)

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 45 sub question (vi)

Answers (1)

Answer: \tan ^{-1} y=\frac{x^{3}}{3}+x+\frac{\pi}{4}

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

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

Solution:

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

        Integrating both sides

        \begin{aligned} &\int \frac{1}{1+y^{2}} d y=\int\left(1+x^{2}\right) d x \\\\ &\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+c \end{aligned}        ...............(1)

        Given that y(0)=1 \text { i.e. at } x=0, y=1

        \begin{aligned} &\tan ^{-1}(1)=0+\frac{0}{3}+c \Rightarrow c=\tan ^{-1}(1)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right) \\\\ &{\left[\therefore \tan \frac{\pi}{4}=1\right]} \\\\ &\Rightarrow c=\frac{\pi}{4} \\\\ &\tan ^{-1} y=x+\frac{x^{3}}{3}+\frac{\pi}{4} \end{aligned}

 

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