Question

Solve the initial-value problem y'+(3/x)y=(3x^4*y^2+10x^2*y+6)/(x^3(2x^2*y+5)), y(1)=1.

Answer 1

@SithsAndGiggles

Answer 2

\[y=\frac{ 3x^4 y^2+10x^2 y+6 }{ x^3 (2x^2 y+5) }-\frac{ 3y }{ x }\]

Answer 3

Wait, supposed to be y' instead of y ^^

Answer 4

\[y'=\frac{ 6-3x^4 y^2-5x^2 y }{ x^3 (2x^2 y+5) }\]

Answer 5

Slight mistake with the above. Missing a factor of \(x\).

This equation looks a lot like the one we did a while back, so I suspect it'll use the same process.
\[\begin{align*}
y'&=\frac{ 6-3x^4 y^2-5x^2 y }{ x^3 (2x^2 y+5) }
\end{align*}\]
Letting \(\color{red}{v=yx}\), you have \(y'=\dfrac{xv'-v}{x^2}\), and the eq. is
\[\begin{align*}
\frac{xv'-v}{x^2}&=\frac{ 6-3v^2-5v }{ x^3 (2v+5) }\\\\
xv'-v&=\frac{ 6-3v^2-5v }{ x(2v+5) }\\\\
xv'&=\frac{ 6-3v^2-5v +vx(2v+5)}{ x (2v+5) }\\\\
v'&=\frac{ 6-3v^2-5v +vx(2v+5)}{ x^2 (2v+5) }\end{align*}\]

Answer 6

Didn't we have the same y'?
\[y=\frac{ 6-3x^4 y^2-5x^2 y }{ x^3 (2x^2 y+5) }\]

Answer 7

Yes, I was referring to the post I deleted. Sorry for the confusion.

Answer 8

So if v=yx, then why do you have \[y=\frac{ 6-3v^2-5v }{ x^3 (2v+5) }\]

Answer 9

Shouldn't v=(x^2)(y)?

Answer 10

Yeah you're right! This is what happens when I don't work out a problem like this on paper. Errors everywhere. Sorry again!

Answer 11

So yeah,
\[v=yx^2~~\implies~~y'=\frac{x^2v'-2xv}{x^4}\]
So the equation should be
\[\begin{align*}
\frac{x^2v'-2xv}{x^4}&=\frac{ 6-3v^2-5v }{ x^3 (2v+5) }\\\\
x^2v'-2xv&=\frac{x( 6-3v^2-5v )}{2v+5 }\\\\
xv'-2v&=\frac{ 6-3v^2-5v }{2v+5 }\\\\
xv'&=\frac{ 6-3v^2-5v +2v(2v+5)}{2v+5 }\\\\
\frac{2v+5}{6+5v+v^2}~dv&=\frac{dx}{x}
\end{align*}\]
a relatively simple separable equation.

Answer 12

Thank you!