# Math Help - Separable Equation Problem.

1. ## 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.

2. ## Re: Separable Equation Problem.

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

use substitutions $u=4y+5$ and $v=2x+3$ to solve integrals .

3. ## Re: Separable Equation Problem.

So is that then integration by parts?

Regards.

4. ## Re: Separable Equation Problem.

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

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

$\int u^{-2}\, du$

5. ## 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.

6. ## Re: Separable Equation Problem.

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 : $du=4dy$ and $dv=2dx$

7. ## 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.

8. ## Re: Separable Equation Problem.

yes , it is..