Need help solving Initial value problem: y' = (3x+2y) / (3x+2y-2) , where y(-1) = -1

[entroqy]

This is as far as I've gotten, I'm stuck lol. Is this seperable? linear? please help!!

 

[MarkFL]

Okay, the original ODE is (accoring to your image, but not the thread title):

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{3x+2y}{3x+2y+2}\)

You used the substitution:

\(\displaystyle \displaystyle u=3x+2y\implies \frac{du}{dx}= 3+2\frac{dy}{dx}\implies \frac{dy}{dx}= \frac{1}{2}\left(\frac{du}{dx}-3\right)\)

And so the ODE becomes:

\(\displaystyle \displaystyle \frac{1}{2}\left(\frac{du}{dx}-3\right)=\frac{u}{u+2}\)

\(\displaystyle \displaystyle \frac{du}{dx}=\frac{2u}{u+2}+3=\frac{5u+6}{u+2}\)

So far so good. I would next write:

\(\displaystyle \displaystyle \frac{du}{dx}=\frac{5u+10-4}{u+2}=\frac{5(u+2)-4}{u+2}=5-\frac{4}{u+2}\)

Can you proceed?

 

[HallsofIvy]

Okay, the original ODE is (accoring to your image, but not the thread title):

\(\displaystyle \displaystyle \frac{dy}{dx}=\frac{3x+2y}{3x+2y+2}\)

You used the substitution:

\(\displaystyle \displaystyle u=3x+2y\implies \frac{du}{dx}= 3+2\frac{dy}{dx}\implies \frac{dy}{dx}= \frac{1}{2}\left(\frac{du}{dx}-3\right)\)

And so the ODE becomes:

\(\displaystyle \displaystyle \frac{1}{2}\left(\frac{du}{dx}-3\right)=\frac{u}{u+2}\)

\(\displaystyle \displaystyle \frac{du}{dx}=\frac{2u}{u+2}+3=\frac{5u+6}{u+2}\)

So far so good.

Yes, excellent.

I would next write:

\(\displaystyle \displaystyle \frac{du}{dx}=\frac{5u+10-4}{u+2}=\frac{5(u+2)-4}{u+2}=5-\frac{4}{u+2}\)

I would not. Instead I would write \(\displaystyle \frac{u+2}{u+ 6}du= \frac{u+ 6- 4}{u+ 6}du= \left(1- \frac{4}{u+6}\right)du= dx\) and integrate both sides.

Can you proceed?

 

[MarkFL]

Haha...hopefully I would have realized my mistake before trying to actually integrate.