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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 23

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 23

Answers (1)

Answer: \tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c

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

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

Solution:

        \begin{aligned} &(1+x)\left(1+y^{2}\right) d x+(1+y)\left(1+x^{2}\right) d y=0 \\\\ &(1+x)\left(1+y^{2}\right) d x=-(1+y)\left(1+x^{2}\right) d y \\\\ &\frac{(1+x)}{1+x^{2}} d x=-\frac{(1+y)}{1+y^{2}} d y \end{aligned}

          Integrating both sides

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

        \begin{aligned} &1+x^{2}=t \Rightarrow 2 x d x=d t \\\\ &1+y^{2}=u \Rightarrow 2 y d y=d u \\\\ &\int \frac{1}{1+x^{2}} d x+\frac{1}{2} \int \frac{1}{t} d t=-\int \frac{1}{1+y^{2}} d y-\frac{1}{2} \int \frac{1}{u} d u \end{aligned}

        \begin{aligned} &\tan ^{-1} x+\frac{1}{2} \log |t|=-\tan ^{-1} y-\frac{1}{2} \log |u|+c \\\\ &\tan ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|=-\tan ^{-1} y-\frac{1}{2} \log \left|1+y^{2}\right|+c \\\\ &\tan ^{-1} x+\tan ^{-1} y+\frac{1}{2} \log \left|1+x^{2}\right|+\frac{1}{2} \log \left|1+y^{2}\right|=c \end{aligned}

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