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  4. The solution curve of (dy/dx)=(y2-2xy-x2/y2+2xy-x2), y(-1)=1 is

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Q. The solution curve of $\frac{dy}{dx}=\frac{y^{2}-2xy-x^{2}}{y^{2}+2xy-x^{2}}$, $y\left(-1\right)=1$ is

Differential Equations

Solution:

Substitute $y=vx $
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, given equation becomes
$\frac{xdv}{dx}+v=\frac{v^{2}-2v-1}{v^{2}+2v-1}$
$\Rightarrow \frac{xdv}{dx}=\frac{v^{2}-2v-1}{v^{2}+2v-1}-v=-\frac{\left(v^{2}+1\right)\left(v+1\right)}{v^{2}+2v-1}$
$\Rightarrow \frac{dx}{x}=\frac{-\left(v^{2}+2v-1\right)dv}{\left(v^{2}+1\right)\left(v+1\right)}=-\left(\frac{2v}{v^{2}+1}-\frac{1}{v+1}\right)dv$
$\Rightarrow ln\,x+ln\,c=ln\left(\frac{v+1}{v^{2}+1}\right)$
$\Rightarrow xc=\frac{v+1}{v^{2}+1}$
$\Rightarrow xc=\frac{x\left(x+y\right)}{x^{2}+y^{2}}$
$\Rightarrow c\left(x^{2}+y^{2}\right)=x+y$
$x=1$,
$y=1$
$\Rightarrow c=0$
$\Rightarrow x+y=0$.