# Thread: need help with two DEs

1. ## need help with two DEs

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

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

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

so

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

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

Then try $y = x u$

Problem 2)

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

so

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

i got $z''=p'p$ and
$p = \frac{z}{1+cz}$
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:
$
z'' = \frac{z'}{1+c_1 z'}
$

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

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

so

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

giving, on integration

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

Solve for $y'$ and then separate.

5. got it, thanks alot, Danny