Let $\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$ be the solution of the differential equation $x \mathrm{~d} y-y \mathrm{~d} x+x y(x \mathrm{~d} y+y \mathrm{~d} x)=0$, $y(1)=2$. Then $\alpha+\beta$ is equal to ________

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Aicharya · Accepted Answer

$\begin{aligned} & a|x|=|y| e^{y x-\beta}, a, b \in N \ & x d y-y d x+x y(x d y+y d x)=0 \ & \frac{d y}{y}-\frac{d x}{x}+(x d y+y d x)=0 \ & \ell n|y|-\ell n|x|+x y=c \ & y(1)=2 \ & \ell n|2|-0+2=c \ & c=2+\ell n 2 \ & \ell n|y|-\ell n|x|+x y=2+\ell n 2 \ & \ell n|x|=\ell n\left|\frac{y}{2}ight|-2+x y \ & |x|=\left|\frac{y}{2}ight| e^{x y-2} \ & 2|x|=|y| e^{x y-2} \ & \alpha=2 \quad \beta=2 \quad \alpha+\beta=4 \end{aligned}$

Let $\alpha|x|=|y| \mathrm{e}^{x y-\beta}, \alpha, \beta \in \mathbf{N}$ be the solution of the differential…

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