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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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 .
So is that then integration by parts? Regards.
Originally Posted by paulbk108 So is that then integration by parts? Regards. No , if you make substitution you should get integral of the form : $\displaystyle \int u^{-2}\, du$
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.
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 $\displaystyle dv=2dx$
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.
yes , it is..
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