# Differential equation

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

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

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

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

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

$
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.
• September 20th 2008, 06:09 AM
shawsend
Quote:

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

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

Have you got any ideas?

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