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## Homework Statement

Solve the following differential equation using seperation of variables

[tex] (1+x)^2 y' = (1-y)^2 , y(1) = 2 [/tex]

## Homework Equations

## The Attempt at a Solution

haven't gotten very far in this at all :/

i've tried dividing both sides by [tex](1+x)^2[/tex], in order to get y' on it's own..

[tex]y' = \frac{(1-y)^2}{(1+x)^2} [/tex]

but i don't know how to integrate this...but had a go anyway

apparently the rule for integrating an expression in brackets is..

[tex] \frac{(ax + b)^n}{a(n+1)} [/tex]

so i tried integrating both halves of the fraction seperatley...giving

[tex]

\frac{(-y+1)^3 }{-3y}[/tex]

[tex]

\frac{(x+1)^2}{3x}

[/tex]

putting these together and dividing gave

[tex]\frac{3x(-y+1)^2}{-3y(x+1)^3} + C[/tex]

however i don't think this is accurage, as substituting in 1 and 2 for x and y respectivley gmade C some out as -3/48. Can someone shed some light at where I've went wrong?

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