Differential Equation: (1+x)^2 dy/dx = (1+y)^2 | Homework Help

[shseo0315]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

So far I'm not getting the right answer, please help.

 

[epenguin]

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!

 

[shseo0315]

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]

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.

 

[mmmboh]

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

 

[shseo0315]

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]

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.