Having trouble getting y by itself after integrating

I'm stuck on two homework problems. The problems are to find the general solution. I'm stuck on the last step which is to get y by itself.

1) The first problem is:

. . .(1 - x^2)dy/dx = 2y

I integrated both sides to get:

. . .1/2ln(y) = ln(x^2 + 1) + C

This is where I'm stuck.

2) The second problem is:

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

I integrated to get:

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

Any help would be great!!!

Same problem:

http://www.mathgoodies.com/forums/topic ... C_ID=30621

Thanks!!! that helped a lot. Can you give me some pointers on the second problem. I had no problem integrating it I'm having trouble getting the answer that's in the back of the book which is y(x)=(1+x)/[1+C(1+x)]. Thanks!!!

For #2:

After integrating, I arrived at \(\displaystyle \frac{-1}{y+1}=\frac{-1}{x+1}+C\), also.

Solving for y:

\(\displaystyle y=\frac{x+Cx+C}{1-CX-C}\)

To be sure, I ran it through the DE solver and received the same answer.

Unless I am messing up somewhere, I get a different answer than the book.

Hello, tegra97!

I got the same answer as Galactus (well, sort of).
I believe your book has a typo . . .

The answer in the back of the book is: \(\displaystyle \L\,y(x)\;=\;\frac{1\,+\,x}{1\,+\,C(1\,+\,x)}\)

Click to expand...

I too got: \(\displaystyle \L\,\frac{-1}{1\,+\,y}\:=\:\frac{-1}{1\,+\,x}\,+\,k\)

Multiply by \(\displaystyle \,-(1\,+\,y)(1\,+\,x):\;\;1\,+\,x\;=\;1\,+\,y \,+\,C(1\,+\,y)(1\,+\,x)\)

Then we have: \(\displaystyle \,1\,+\,x\;=\;1\,+\,y\,+\,C\,+\,Cx\,+\,Cy\,+\,Cxy\)

Rearrange terms: \(\displaystyle \,y\,+\,Cy\,+\,Cxy\;=\;x\,-\,C\,-\,Cx\)

Factor: \(\displaystyle \,y(1\,+\,C\,+\,Cx)\;=\;x\,-\,C\,-\,Cx\)

Therefore: \(\displaystyle \L\,y\;=\;\frac{x\,-\,C\,-\,Cx}{1\,+\,C\,+\,Cx} \;= \;\frac{x\,-\,C(1\,+\,x)}{1\,+\,C(1\,+\,x)}\)

Thanks for going over the steps Soroban. I had trouble seeing that you had to multiply by -(1+y)(1+x). Yeah, it's probably a typo because our professor said there are some typos in this book.