Let |x|=|y|e^xy- , , N be the solution of the differential equation xdy-ydx+xy(xdy+ydx)=0,…

Question

Let $\alpha |x|=|y|{e}^{xy-\beta },\alpha ,\beta \in N$ be the solution of the differential equation $xdy-ydx+xy(xdy+ydx)=0,y(1)=2$. Then $\alpha +\beta $ is equal to \_\_\_\_\_\_\_\_

Answer: 4

Accepted Answer

$\begin{gathered} \alpha |x|=|y|{e}^{xy-\beta } \ \frac{xdy-ydx}{{y}^{2}}+\frac{xy(xdy+ydx)}{{y}^{2}}=0 \ -d(\frac{x}{y})+\frac{x}{y}d(xy)=0 \end{gathered}$

$\begin{gathered} \int d(xy)=\int \frac{d(\frac{x}{y})}{\frac{x}{y}} \ xy=\ln |\frac{x}{y}|+\ln c \ xy=\ln (|\frac{x}{y}|\cdot c) \ \because y(1)=2 \ 2=\ln |\frac{1}{2}|c\Rightarrow c=2{e}^{2} \ 	herefore \, solutionxy=\ln (|\frac{x}{y}|\cdot 2{e}^{2}) \ {e}^{xy}=\frac{|x|}{|y|}\cdot 2{e}^{2} \ 2|x|=|y|{e}^{xy-2} \ \Rightarrow \alpha =2,\beta =2,\alpha +\beta =4 \end{gathered}$

Solutions · 1