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?
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.
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:
$\displaystyle \frac{4y}{4-3y}\;dy=dx$
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: $\displaystyle \frac{4}{3}(4-3y)-\frac{8}{3}\ln{|4-3y|}=x+C$