<s>[INST] Nonlinear first order differential equation (exact) 
Find an implicit solution to the initial value problem $$(x+y)^2+(2xy+x^2-1)\dfrac{dy}{dx} = 0 \\ y(1) = 1$$

Attempted solution
The equation can be written as  $M(x,y)+N(x,y)y'=0$. We try to find find the potential function $\psi(x,y)$: $$\psi_x = M(x,y) = (x+y)^2 \\ \psi_y = N(x,y) = 2xy+x^2-1 $$
Integrating last eq. wrt $y$ yields $\psi(x,y) = xy^2+x^2y-y+f(x)$, so that $M(x,y)=\psi_x \Leftrightarrow y^2+2xy+x^2= y^2+2xy+f'(x) \Rightarrow f'(x) = x^2 \Rightarrow f(x) = \dfrac{x^3}{3}+C $
Okay so now we have the potential $\psi(x,y) = xy^2+x^2y-y + \dfrac{x^3}{3}+C$. 
The problem
Plugging in $y(1)=1$ yields $\psi(1,1) = \dfrac{4}{3}+C$. What should I set this equal to?
 [/INST] Conventionally an implicit solution is of the form $\psi(x,y) =0$. To this end, you should set $$C=-\frac{3}{4}.$$
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