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The particular solution of differential equation $(\mathrm{x}+\mathrm{y})…
The particular solution of differential equation $(\mathrm{x}+\mathrm{y}) \mathrm{dy}+(\mathrm{x}-\mathrm{y}) \mathrm{d} \mathrm{x}=0$ at $\mathrm{x}=\mathrm{y}=1$ is
$\log \left|\frac{\mathrm{x}^2+\mathrm{y}^2}{2}\right|=\frac{\pi}{2}-2 \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ $\log \left|\mathrm{x}^2+\mathrm{y}^2\right|=\frac{\pi}{2}-2 \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ $\log \left|\frac{\mathrm{x}^2+\mathrm{y}^2}{2}\right|=\frac{\pi}{4}-\tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ $\log \left|\mathrm{x}^2+\mathrm{y}^2\right|=\frac{\pi}{4}-2 \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$ Solution
$(x+y) d y+(x-y) d x=0$
We have $\frac{d y}{d x}=-\frac{(x-y)}{x+y}$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore v+x \frac{d v}{d x}=\frac{-(x-v x)}{x+v x}=\frac{-x(1-v)}{x(1+v)}=\frac{-(1-v)}{1+v}$
$\therefore x \frac{d v}{d x}=\frac{-1+v}{1+v}-v=\frac{-1+v-v-v^2}{1+v} \Rightarrow x \frac{d v}{d x}=\frac{-\left(1+v^2\right)}{1+v}$
$\therefore \int \frac{(1+v)}{1+v^2} d v=-\int \frac{d x}{x} \Rightarrow \int \frac{d v}{1+v^2}+\frac{1}{2} \int \frac{2 v}{1+v^2} d v=-\int \frac{d x}{x}$
$\begin{aligned}
& \therefore \tan ^{-1}(\mathrm{v})+\frac{1}{2} \log \left|1+\mathrm{v}^2\right|=-\log \mathrm{x}+\log \mathrm{c}_1 \\
& \quad \Rightarrow \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)+\frac{1}{2} \log \left|1+\frac{\mathrm{y}^2}{\mathrm{x}^2}\right|=-\log \mathrm{x}+\log \mathrm{c}_1
\end{aligned}$
$2 \tan ^{-1}\left(\frac{y}{x}\right)+\log \left|\frac{x^2+y^2}{x^2}\right|+2 \log x=\log c \ldots\left[\because \log c=2 \log c_1\right]$
$\therefore 2 \tan ^{-1}\left(\frac{y}{x}\right)+\log \left|\frac{x^2+y^2}{x^2} \times x^2\right|=\log c$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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