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.

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- Dec 1st 2011, 11:51 PMpaulbk108Separable 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 2nd 2011, 12:06 AMprincepsRe: Separable Equation Problem.
- Dec 2nd 2011, 12:34 AMpaulbk108Re: Separable Equation Problem.
So is that then integration by parts?

Regards. - Dec 2nd 2011, 12:40 AMprincepsRe: Separable Equation Problem.
- Dec 2nd 2011, 12:51 AMpaulbk108Re: 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, 01:17 AMprincepsRe: Separable Equation Problem.
- Dec 2nd 2011, 01:59 AMpaulbk108Re: 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, 02:19 AMprincepsRe: Separable Equation Problem.
yes , it is..