Differential equation

  • Thread starter shseo0315
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  • #1
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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.
 

Answers and Replies

  • #2
epenguin
Homework Helper
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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!
 
  • #3
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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.
 
  • #4
Dick
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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.
 
  • #5
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Inverse both sides to find y+1, and then just subtract one from both sides to solve for y.
 
  • #6
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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
 
  • #7
Dick
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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.
 

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