# Thread: need help with two DEs

1. ## need help with two DEs

can't separate anything from this DE :
$\displaystyle (xy'-y)^2=y'^2-2yy'/x+1$
tried to group something but it didn't help
$\displaystyle x^2y'(xy'-2y)+xy^2=y'(xy'-2y)+x$
and the second problem is this below

2. Originally Posted by noteiler
can't separate anything from this DE :
$\displaystyle (xy'-y)^2=y'^2-2yy'/x+1$
tried to group something but it didn't help
$\displaystyle x^2y'(xy'-2y)+xy^2=y'(xy'-2y)+x$
and the second problem is this below
For problem 1) Multiply by $\displaystyle x^2$ so

$\displaystyle x^2 \left(x y' - y\right)^2 = x^2 y'^2 - 2 x y' y + x^2$

so

$\displaystyle x^2 \left(x y' - y\right)^2 = \left(x y' - y\right)^2 + x^2 - y^2$

or $\displaystyle x y' - y = \pm \frac{\sqrt{x^2-y^2}}{\sqrt{x^2-1}}$

Then try $\displaystyle y = x u$

Problem 2)

$\displaystyle z'' = \frac{z'}{1+c_1 z'}$

so

$\displaystyle \left(1+c_1z'\right)z'' = z'$

Integrate left and right sides once. See how that goes.

3. thanks alot for 1. answer, but i didn't really understand how that becomes:
$\displaystyle z'' = \frac{z'}{1+c_1 z'}$

i got $\displaystyle z''=p'p$ and
$\displaystyle p = \frac{z}{1+cz}$
but how can i get it from thease?

4. Originally Posted by noteiler
thanks alot for 1. answer, but i didn't really understand how that becomes:
$\displaystyle z'' = \frac{z'}{1+c_1 z'}$

i got $\displaystyle z''=p'p$ and
$\displaystyle p = \frac{z}{1+cz}$
but how can i get it from thease?
Sorry, my bad. What I should of said is

$\displaystyle y'' = \frac{y'}{1+c_1 y'}$

so

$\displaystyle (1+c_1 y') y'' = y'$

giving, on integration

$\displaystyle y' + \frac{c_1}{2} y'^2 = y + c_2$.

Solve for $\displaystyle y'$ and then separate.

5. got it, thanks alot, Danny