Im trying to solve this D.E:
$$(x^2-xy+y^2)dx+(-xy)dy=0, y(1)=0$$
After I subsituted $$y=ux$$ $$dy=xdu+udx$$
I get $$-ln(x)=ln(1-u)-1+u+c$$
Answer: $$-xln(x-y)=y$$
assume () are absolute value
I truly appreciate everyone's help. I really do learn a lot from this website.
After the substitution we get
\begin{align*} &&x^2(1-u+u^2)dx-ux^2(xdu+udx)&=0\\[3pt] \iff &&x^2(1-u)dx-ux^3du&=0 \\[3pt] \implies &&\frac{dx}{x}+\left(1-\frac{1}{1-u}\right)du&=0\qquad x(1-u)\neq 0\\[3pt] \implies &&\ln|x|+u-\ln|1-u|&=c\\[3pt] \implies &&x\left(1-u\right)&=\pm e^{c-u}\\[3pt] \implies &&x-y&=Ce^{-y/x}\qquad\text{where }C=\pm e^{-c}\\[3pt] \end{align*} From the initial condition we find $C=1$, and then $$\boxed{\color{blue}{x-y=e^{-y/x}}}$$ Which is an implicit solution.