1. ## Help =)

Having problems solving this diff equation.

(1-x^2)y´+ xy = x for -1 < x < 1

I get the ínt factor to 1/(1-x^2)^0,5, But having problems after that.

Thank you.

2. Originally Posted by echo123
Having problems solving this diff equation.

(1-x^2)y´+ xy = x for -1 < x < 1

I get the ínt factor to 1/(1-x^2)^0,5, But having problems after that.

Thank you.
does post #21 here help?

3. Hello again

Not really,,I understand the general approach to solving a diff equation its just that I cant get the right side to be a single derivate. having problems there.

4. Im sorry, I meant the left side =)

5. Originally Posted by echo123
Hello again

Not really,,I understand the general approach to solving a diff equation its just that I cant get the right side to be a single derivate. having problems there.
ok, so following the steps exactly, you would end up with

$\displaystyle \frac 1{\sqrt{1 - x^2}}y' + \frac x{(1 - x^2)^{3/2}}y = \frac x{(1 - x^2)^{3/2}}$

again, just following the steps i outlined (we literally don't have to think about this if we know the steps by heart), this becomes

$\displaystyle \left( \frac 1{\sqrt{1 - x^2}}y\right)' = \frac x{(1 - x^2)^{3/2}}$

now integrate both sides and continue

look closely at the steps and see what is confusing you. the single derivative is simply the coefficient of y' times y.

6. Oh,, I did exatly that but I thought it was wrong. Thank you very much.

7. Originally Posted by echo123
Oh,, I did exatly that but I thought it was wrong. Thank you very much.
you're welcome

next time have more confidence in yourself and tell us what you did, so we can give you a pat on the back

8. I messed up when I multiplied the coefficient of y' and y. I Do that to check that im on the right track but this time it turned out bad =(.