# Differential equation

• Sep 20th 2008, 05:15 AM
Aliquantus
Differential equation
Hi, I'm having problems with the following differential equation:

$\displaystyle \frac{dy}{dx}=\frac{1}{y}-\frac{3}{4}$

Have you got any ideas?
• Sep 20th 2008, 06:01 AM
Peritus
sure, I have a pretty good idea.
first you rearrange the equation like so:

$\displaystyle \frac{{4y}} {{4 - 3y}}dy = dx$

so this is basically a separable equation, all you have to do now is to integrate both sides.

$\displaystyle y + \frac{4} {3}\ln \left| {4 - 3y} \right| = - \frac{3} {4}x + C$

I doubt that an explicit form of the solution can be obtained.
• Sep 20th 2008, 06:09 AM
shawsend
Quote:

Originally Posted by Aliquantus
Hi, I'm having problems with the following differential equation:

$\displaystyle \frac{dy}{dx}=\frac{1}{y}-\frac{3}{4}$

Have you got any ideas?

$\displaystyle \frac{4ydy}{4-3y}=dx$
$\displaystyle \frac{4y}{4-3y}\;dy=dx$
Then you integrate both sides: $\displaystyle 4\int\frac{y}{4-3y}\;dy=\int dx$
Make the substitution $\displaystyle u=4-3y$...rest follows
Should end up with this: $\displaystyle \frac{4}{3}(4-3y)-\frac{8}{3}\ln{|4-3y|}=x+C$