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Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 33Maths Textbook Solution.

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Answer: \mathrm{e}^{y}+\mathrm{e}^{-y}-\log |\mathrm{x}|+\frac{\mathrm{x}^{2}}{2}+\mathrm{C}

Hint: you must know the rules of solving differential equation and integrations.

Given:x\left ( e^{2y}-1 \right )dy+\left ( x^{2}-1 \right )e^{y}dx=0

Solution:x\left ( e^{2y}-1 \right )dy+\left ( x^{2}-1 \right )e^{y}dx=0

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

Integrating both sides,

\int\left(e^{y}-e^{-y}\right) d y=\int \frac{1}{x} d x-\int x d x \\ \int\left(e^{y} d y-\int e^{-y} d y=\int \frac{1}{x} d x-\int x d x\right. \\

e^{y}-\left(-e^{-y}\right)=\log |x|-\frac{x^{2}}{2}+c

\Rightarrow e^{y}+e^{-y}-\log |x|+\frac{x^{2}}{2}+c

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