I've tried unsuccessfully to solve these two problems. I'd grateful for any help here.
Differential Equation: $y' = (x^2+2xy-y^2)/(x^2-xy)$
Here's what I did: $$f(\lambda x,\lambda y) = (\lambda ^2x^2+2\lambda ^2xy-\lambda y^2)/(\lambda ^2x^2-\lambda ^2xy) \\ = (\lambda ^2(x^2+2xy-y^2))/(\lambda ^2(x^2-xy)) \\= (x^2+2xy-y^2)/(x^2-xy) = f(x,y)$$
And I don't know what to do next.
Cauchy Equation: $4(2y^3+xy-y)y' = 1, y(0) = 0$
With this one, I tried moving everything but $y'$ to the right, I also tried moving $4$ to the right and I also tried solving this equation without moving anything and it just doesn't work for me.
I hope you'll be able to help.
1) You proved that the equation is homogeneous $$y' = \frac {(x^2+2xy-y^2)}{(x^2-xy)}$$ Substitute $y=tx \implies y'=t'x+t$ $$t'x+t=\frac {1+2t-t^2}{1-t}$$ $$t'x=\frac {1+t}{1-t}$$ This equation is separable
$$\int \frac {1-t}{1+t}dt=\int \frac {dx}{x}$$
2) $$4(2y^3+xy-y)y' = 1, y(0) = 0$$ Consider $x'$ instead of $y'$ $$x'=4(2y^3+xy-y) $$ $$x'-4xy=4(2y^3-y) $$ This equation is a first order linear equation
Use integrating factor or other techniques