Solve the given differential equation.

$y\prime = \dfrac{(3x^2-1)}{3+2y)}$

Where would I begin?

Question

Solve the given differential equation.

$y\prime = \dfrac{(3x^2-1)}{3+2y)}$

Where would I begin?

anonymous · Answer

As simple as this:$$y'=\frac{ dy }{ dx }=\frac{ 3x^2-1 }{ 3+2y } \rightarrow (3+2y)dy=(3x^2-1)dx$$You can integrate

austinl · Answer

Oh my goodness. That makes gobs more sense.

austinl · Answer

Ok, so now I have:

$y^2+3y+C=x^3-x+C$

Now, I haven't a clue what to do next, combine like terms and solve for y?

austinl · Answer

$\large{y=\pm \dfrac{-(\sqrt{4x^3-4x+9}+3)}{2}}$

That makes absolutely zero sense.

austinl · Answer

+C on the end of course....

loser66 · Answer

lol!!! Actually, you don't need to find y out, just let it there as  y ^2 +3y -x^3 +x = C
that's it

austinl · Answer

Many typos there.

$y^2+3y-x^3+x=C~,~y\ne\frac{-3}{2}$

loser66 · Answer

it's ok, looks good to me

austinl · Answer

Woohoo, very nice!

loser66 · Answer

hahahaha... glad to see "Woohoo..." wudwud... You are gud

anonymous · Answer

If you leave the solution like: $$3y+y^2-x^3+x=C$$is perfectly valid