I am having problems solving this differential equation:

(dy/dx) = (x + y)^2

Any help would be appreciated.

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- Jan 31st 2007, 10:51 PMmachi4velliProblem with a Diff Equation
I am having problems solving this differential equation:

(dy/dx) = (x + y)^2

Any help would be appreciated. - Feb 1st 2007, 05:15 AMThePerfectHacker
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.