I am having problems solving this differential equation:
(dy/dx) = (x + y)^2
Any help would be appreciated.
We have,
$\displaystyle y'=(x+y)^2$
Define a new function,
$\displaystyle z=x+y$.
Then,
$\displaystyle z'=1+y'\to y'=z'-1$
Thus,
$\displaystyle z'-1=z^2$
$\displaystyle z'=1+z^2$
$\displaystyle \frac{z'}{1+z^2}=1$
$\displaystyle \int \frac{z'}{1+z^2} dx=\int 1 dx$
$\displaystyle \tan^{-1} z=x+C$
$\displaystyle \tan^{-1} (x+y)=x+C$
Are all the solutions on some non-zero open interval.