1. ## Separable Equations

Solve the equation
$\displaystyle x dy/dx = y^2 - y$
$\displaystyle 1/x dx = 1/(y^2 -y) dy$
$\displaystyle ln x + C = ln (y^2-y)$
$\displaystyle e^l^n^x * e^C = y^2 - y$
$\displaystyle Cx = y^2 - y$
$\displaystyle y^2 - y -Cx = 0$
I'm pretty sure i did something wrong with this one

Solve the equation
$\displaystyle e^xy dy/dx = e^-^y + e^-^2^x^-^y$
No idea where to start on this so any help is greatly appreciated

2. I'm not really sure what you did, after separation you should have $\displaystyle \displaystyle \frac{dy}{y^2-y}= \frac{dx}{x}$

3. Originally Posted by Naples
Solve the equation
$\displaystyle x dy/dx = y^2 - y$
$\displaystyle 1/x dy/dx = 1/(y^2 -y) dy$
$\displaystyle ln x + C = ln (y^2-y)$
$\displaystyle e^l^n^x * e^C = y^2 - y$
$\displaystyle Cx = y^2 - y$
$\displaystyle y^2 - y -Cx = 0$
I'm pretty sure i did something wrong with this one

Solve the equation
$\displaystyle \displaystyle e^xy dy/dx = e^-^y + e^-^2^x^-^y$
No idea where to start on this so any help is greatly appreciated
$\displaystyle \displaystyle x\frac{dy}{dx}=y^2-y \iff \frac{dy}{y(y-1)}=\frac{dx}{x}$

Now by partial fractions or other algebraic manipulations we get

$\displaystyle \displaystyle \left(\frac{-1}{y}+\frac{1}{y-1}\right)dy=\frac{dx}{x}$

Now we can integrate both sides

For the 2nd one note that

$\displaystyle \displaystyle e^{-y}+e^{-2x-y}=e^{-y}(1+e^{-2x})$

So once again you can separate the equation.
Can you finish from here?

4. $\displaystyle e^xy dy/dx = e^-^y(1+ e^-^2^x)$
$\displaystyle ye^ydy = (e^-x + e^-^3^x)dx$
using integration by parts
$\displaystyle ye^y - e^y = -e^-^x - (1/3)e^-^3^x + C$
$\displaystyle e^y(y -1) = -e^-^x - (1/3)e^-^3^x + C$
Hopefully I did this right, if so would it be necessary to simplify this further?