Question

Solve the given differential equation:

$y^{2}dx+(x^2-xy+y^2)dy=0$

Answer 1

First check weather it is homogeneous or not!
For checking we rearrange the equation

$y^{2}dx+(x^2-xy+y^2)dy=0$

$\frac{dy}{dx}=\frac{-y^2}{x^2-xy+y^2}$ ....... $\left(1\right)$

Now, when we put $y=vx$ in right side then found that only function of $v$

So, we can say that it is homogeneous equation.

So, put $y=vx$ $\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation $\left(1\right)$, we get

$v+x\frac{dv}{dx}=\frac{-v^2}{1-v+v^2}$

$x\frac{dv}{dx}=\frac{-v^2}{1-v+v^2}-v$

$x\frac{dv}{dx}=\frac{-v-v^3}{1-v+v^2}$

$\frac{(1+v^2)-v}{-v(1+v^2)}dv=\frac{dx}{x}$

$\frac{-1}{v}dv+\frac{1}{1+v^2}dv=\frac{dx}{x}$

Now, integrate both sides

$-\ln v+{\tan }^{-1}v=\ln x+\ln c$ where c=contant

Now: ${\tan }^{-1}v=\ln x+\ln v+\ln c$

${\tan }^{-1}v=\ln \left(xcv\right)$

$\Rightarrow e^{{\tan }^{-1}v}=xcv$

Resubstitute $v=\frac{y}{x}$

$\Rightarrow e^{{\tan }^{-1}\frac{y}{x}}=xc\frac{y}{x}$

$\Rightarrow e^{{\tan }^{-1}\frac{y}{x}}=yc$

where c=contant