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Please solve RD  Sharma class 12 Chapter 21 Difrential Equations  excercise 21.6 question 1 maths textbook solution

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

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

Hint: Separate y and x and then integrate both sides.


\begin{aligned} &\begin{aligned} \frac{d y}{d x}+\frac{1+y^{2}}{y} &=0 \\ \frac{d y}{d x} &=-\left(\frac{1+y^{2}}{y}\right) \end{aligned}\\ &\left(\frac{y}{1+y^{2}}\right) d y=-d x\\ &\text { [integrate both side] }\\ &\int\left(\frac{y}{1+y^{2}}\right) d y=-\int d x \end{aligned}

\begin{aligned} &\text { Let } 1+y^{2}=t\\ &\text { [differentiate with respect to } y \text { ] }\\ &\begin{aligned} 2 y d y &=d t \\ & \frac{1}{2} \int \frac{1}{t} d t=-\int d x \\ & \frac{1}{2} \log |t|=-x+C \\ & \frac{1}{2} \log \left|1+y^{2}\right|+x=C \end{aligned} \end{aligned}


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