# Differential equation

## Homework Statement

(1+x)^2 dy/dx = (1+y)^2

## The Attempt at a Solution

The post I put up a while ago actually turns out to be the one above.

epenguin
Homework Helper
Gold Member
You can get everything involving x on one side, everything involving y on the other. Called 'variables separable'. Look up.

Edit - just as was your previous one I have now seen which you did manage to do!

Thanks.
But on the way, 1/(1+x^2)dx = 1/(1+y^2)dy
if I take an intergral, I get

-1/(x+1) + c = -1/(y+1) + c

This is 1/(x+1) + c = 1/(y+1) right?

The answer states that y = (1+x)/[1+c(1+x)] -1

I don't know how to get there.

Dick
Homework Helper
Thanks.
But on the way, 1/(1+x^2)dx = 1/(1+y^2)dy
if I take an intergral, I get

-1/(x+1) + c = -1/(y+1) + c

This is 1/(x+1) + c = 1/(y+1) right?

The answer states that y = (1+x)/[1+c(1+x)] -1

I don't know how to get there.

Right. You never really needed two c's. Just take your expression and use algebra to solve for y.

Inverse both sides to find y+1, and then just subtract one from both sides to solve for y.

Right. You never really needed two c's. Just take your expression and use algebra to solve for y.

from 1/(x+1) +c = 1/(y+1)

that is y+1+c = x+1 right

y(x) = x+c

this is what I get, but the answer is quite different

which is y = (1+x)/[1+c(1+x)] -1

Dick
Homework Helper
from 1/(x+1) +c = 1/(y+1)

that is y+1+c = x+1 right

y(x) = x+c

this is what I get, but the answer is quite different

which is y = (1+x)/[1+c(1+x)] -1

1 over 1/(x+1)+c isn't equal to (x+1)+c. Use correct algebra. Not just any algebra.