The solution of the differential equation (x^2+y^2)dx-5xydy=0,y(1)=0, is :

Question

The solution of the differential equation $({x}^{2}+{y}^{2})dx-5xydy=0,y(1)=0$, is :

Accepted Answer

$\begin{gathered} ({x}^{2}+{y}^{2})dx-5xydy=0 \ \frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}}{5xy} \ y=vx \ \frac{dy}{dx}=v+x\frac{dv}{dx} \ V+x\frac{dv}{dx}=\frac{1+{v}^{2}}{5v} \ x\frac{dv}{dx}=\frac{1+{v}^{2}-5{v}^{2}}{5v} \end{gathered}$

$\begin{gathered} \frac{1}{8}\int \frac{8	imes 5vdv}{1-4{v}^{2}}=\int \frac{dx}{x} \ \frac{-5}{8}\ln |1-4{v}^{2}|=\ln |x|+\ln c \ \Rightarrow \frac{5}{8}\ln \frac{|{x}^{2}-4{y}^{2}|}{{x}^{2}}+\ln |x|=\ln c \ {|\frac{{x}^{2}-4{y}^{2}}{{x}^{2}}|}^{5/8}|x|=c \ \frac{{|{x}^{2}-4{y}^{2}|}^{5/8}}{|x{|}^{\frac{1}{4}}}=c\because y(1)=0\Rightarrow c=1 \end{gathered}$

$\begin{gathered} \Rightarrow {|{x}^{2}-4{y}^{2}|}^{5/8}=|x{|}^{\frac{1}{4}} \ {|{x}^{2}-4{y}^{2}|}^{5}={x}^{2} \end{gathered}$

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