# Differential equation

• Sep 20th 2008, 06:15 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?
• Sep 20th 2008, 07:01 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.
• Sep 20th 2008, 07:09 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?

How about separating variables:

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

Hey, I think that turns into an interesting problem: Express the solution as y=f(x).
• Sep 20th 2008, 07:13 AM
polymerase
You have to seperate the two variables such that all the y's are on one side and all the x's are on the other....so use cross multiplication...you should end up with this:
$\frac{4y}{4-3y}\;dy=dx$
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: $\frac{4}{3}(4-3y)-\frac{8}{3}\ln{|4-3y|}=x+C$