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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.

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$$x^2 + 2xy -y^2 = (2xy - x^2 + e^y)y'$$ $$(x^2 + 2xy -y^2)dx -(2xy - x^2 + e^y)dy=0$$ 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.

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  • $\begingroup$ Thanks for the help, however, i'm just a bit unsure about how you did the rearranging. For example, how did $(2xydx+x^2 dy)$ become $d(x^2 y)$ $\endgroup$ – N_Mathematics_B Mar 12 '20 at 6:02
  • $\begingroup$ Why ? what makes you unsure ? @N_Mathematics_B Note that you can solve it with exactness as shown in the link I provided. $\endgroup$ – Aryadeva Mar 12 '20 at 6:03
  • $\begingroup$ Im just unsure how $(2xydx+x^2 dy)$ became $d(x^2 y)$ in the rearranging process. $\endgroup$ – N_Mathematics_B Mar 12 '20 at 6:08
  • $\begingroup$ This has nothing to do with rearranging Try to differentiate $x^2y$ that will help you @N_Mathematics_B $\endgroup$ – Aryadeva Mar 12 '20 at 6:10
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    $\begingroup$ Ohhh, I understand, thanks for all the help $\endgroup$ – N_Mathematics_B Mar 12 '20 at 6:12

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