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I am trying to solve the differential equation $\frac{dy}{dx}=(x+y)^{2}$, $y(0)=0$. What I have tried so far is the substitution $y(x)=-x+z(x)$ and $z(0)=0$, this simplifies the differential equation to:

$$\frac{dz}{dx}=1+z^{2}$$

Which I can solve by separation of variables to give $z(x)=\tan{x}$, this gives $y(x)=-x+\tan{x}$. This is a solution to the differential equation in the title if I test it with the given differential equation, but it is not continuous in $\pi/2+k\pi,k\in\mathbb{Z}$. So I believe I am doing something wrong as I expect the solution to be continuous. Is my solution right? Or should I be approaching this differently?

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  • $\begingroup$ Why do you expect the solution to be continuous? $\endgroup$ Feb 5, 2014 at 17:58

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By separation of variables you find a family of integral curves $\arctan{(y+x)}=x+C$. The choice of an integral curve passing through the point $(x,y)=(0,0)$ implies that $C=0$, i.e., solution has been found in the implicit form $\arctan{(y+x)}-x=0$. To redefine this solution explicitly, notice that along the required integral curve $x\in \bigl(-\frac{\pi}{2},\frac{\pi}{2}\bigr)$. Hence, the explicit form $\,y(x)=-x+\tan{x}\,$ may be valid only for $x\in \bigl(-\frac{\pi}{2},\frac{\pi}{2}\bigr)$.

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