Solve the initial value problem:
$x^2 + 2xy -y^2 = (2xy - x^2 + e^y)y'$
Where $y(1)=1/2$
Answer the question with the relation y as a function of x.
I've been working on the following question and can't quite seem to figure out what to do. I understand that this is a Non-linear first-order ODE, and I think that it might be separable. The thing is I don't quite know how to solve this if it is separable. I thought that maybe the first step is to divide, so it becomes:
$(x^2 + 2xy -y^2)\div(2xy - x^2 + e^y) = y'$
I'm not quite sure how to get it in the following form: $dy/dx = f(x)g(y)$ (form for seperable ODE)
Any help with this would be much appreciated.
$$x^2 + 2xy -y^2 = (2xy - x^2 + e^y)y'$$ $$(x^2 + 2xy -y^2)dx -(2xy - x^2 + e^y)dy=0$$ It's exact, we have: $$Mdx+Ndy=0 \implies \partial_y M=\partial_x N$$ You can solve this DE with exactness. Look here : Exact Differential
Another way: $$(x^2 + 2xy -y^2)dx -(2xy - x^2 + e^y)dy=0$$ Rearrange terms: $$x^2dx + (2xydx+x^2dy) -(y^2dx +2xydy)-e^ydy=0$$ $$\frac 13dx^3 + d(x^2y) -d(y^2x)-de^y=0$$ After integration: $$\frac 13x^3 + x^2y -y^2x-e^y=C$$ To find $C$ apply initial condition.