# Separable Equation Problem.

• Dec 1st 2011, 10:51 PM
paulbk108
Separable Equation Problem.
Having not studied calculus for a couple of years I am a tad rusty. Was hoping someone could give me a kick start.

The question I need to solve is:

dy/dx = [(4y + 5)/(2x + 3)]^2

Any help would be much appreciated...

Regards.
• Dec 1st 2011, 11:06 PM
princeps
Re: Separable Equation Problem.
Quote:

Originally Posted by paulbk108
Having not studied calculus for a couple of years I am a tad rusty. Was hoping someone could give me a kick start.

The question I need to solve is:

dy/dx = [(4y + 5)/(2x + 3)]^2

Any help would be much appreciated...

Regards.

$\displaystyle \frac{dy}{dx}=\frac{(4y+5)^2}{(2x+3)^2} \Rightarrow \int \frac{1}{(4y+5)^2} \, dy=\int \frac{1}{(2x+3)^2} \, dx$

use substitutions$\displaystyle u=4y+5$ and $\displaystyle v=2x+3$ to solve integrals .
• Dec 1st 2011, 11:34 PM
paulbk108
Re: Separable Equation Problem.
So is that then integration by parts?

Regards.
• Dec 1st 2011, 11:40 PM
princeps
Re: Separable Equation Problem.
Quote:

Originally Posted by paulbk108
So is that then integration by parts?

Regards.

No , if you make substitution you should get integral of the form :

$\displaystyle \int u^{-2}\, du$
• Dec 1st 2011, 11:51 PM
paulbk108
Re: Separable Equation Problem.
So;

int [u^-2} = int [v^-2]

= (u^-1)/-1 = (v^-1)/-1

= [(4y + 5)^-1]/-1 = [(2x + 3)^-1]/-1 + c

Doesn't seem correct...?

Regards.
• Dec 2nd 2011, 12:17 AM
princeps
Re: Separable Equation Problem.
Quote:

Originally Posted by paulbk108
So;

int [u^-2} = int [v^-2]

= (u^-1)/-1 = (v^-1)/-1

= [(4y + 5)^-1]/-1 = [(2x + 3)^-1]/-1 + c

Doesn't seem correct...?

Regards.

Note that :$\displaystyle du=4dy$ and $\displaystyle dv=2dx$
• Dec 2nd 2011, 12:59 AM
paulbk108
Re: Separable Equation Problem.
u = 4y + 5
=> du/dy = 4
=>dy = du/4

=> int [dy/(4y + 5^2)] = 1/4 int [du/u^2] = ...?

v = 2x + 3
=>dv/dx = 2
=>dx = dv/2

=> int [dx/(2x + 3)^2] = 1/2 int [dv/v^2] = ...?

Is that correct?

Regards.
• Dec 2nd 2011, 01:19 AM
princeps
Re: Separable Equation Problem.
yes , it is..