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Please solve RD  Sharma class 12 Chapter 21 Difrential Equations  excercise 21.6 question 2 maths textbook solution

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

Given: \frac{d y}{d x}=\frac{1+y^{2}}{y^{3}}

Hint: Use integration by parts method 

Solution:  \frac{d y}{d x}=\frac{1+y^{2}}{y^{3}}

               \frac{y^{3}}{1+y^{2}} d y=d x

On dividing  \frac{y^{3}}{1+y^{2}} we will write it as y-\frac{y}{y^{2}+1} \text { i.e., } \mathrm{Q}+\frac{R}{D}


\int\left(y-\frac{y}{y^{2}+1}\right) d y=\int d x \; \; \; \; \; \; \; \; [integrate \: both\: sides]

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

\text { Let } y^{2}+1=t                    [ Diffrentiate with reference to y]

\begin{aligned} & 2 y d y=d t \\ &\qquad \begin{array}{c} \frac{y^{2}}{2}-\frac{1}{2} \int \frac{1}{t} d t=x \\ \\\frac{y^{2}}{2}-\frac{1}{2} \log |t|=x+C \end{array} \\ \\&\qquad \begin{array}{l} \frac{y^{2}}{2}-\frac{1}{2} \log \left|y^{2}+1\right|=x+C \end{array} \end{aligned}


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