Question

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

Answer 1

@SithsAndGiggles

Answer 2

I'll come back in 25 minutes.

Answer 3

I can see that the equation can be treated with the exact method, but do you have a suggested substitution anywhere?

Answer 4

No. But I'd like to use the substitution method. So what's the substitution for y?

Answer 5

I haven't found a useful substitution yet... The exact-equation method works well enough, but if you haven't learned it yet you shouldn't be expected to use it.

Answer 6

This is the last couple of problems before the exact equation method. So I shouldn't use the exact-method.

Answer 7

I had an idea earlier that would involve completing the square, but I'm not so sure it's useful.

Another idea was to move the \(\dfrac{2}{x}y\) term to the right side. The numerator simplifies a bit, but the step or sub after that is unclear.

Answer 8

The answer is...\[y=\frac{ -3+\sqrt{1+60x} }{ 2x }\]

Answer 9

Hold on, let's see if there's any merit to my suggestions:
\[\begin{align*}
y'+\frac{2y}{x}&=\frac{3x^2y^2+6xy+2}{x^2(2xy+3)}
\end{align*}\]
"Simplifying" method:
\[\begin{align*}
y'&=\frac{3x^2y^2+6xy+2}{x^2(2xy+3)}-\frac{2y}{x}\\\\
y'&=\frac{3x^2y^2+6xy+2-2xy(2xy+3)}{x^2(2xy+3)}\\\\
y'&=\frac{2-x^2y^2}{x^2(2xy+3)}
\end{align*}\]
Completing the square:
\[\begin{align*}
y'+\frac{2y}{x}&=\frac{3x^2y^2+6xy+2}{x^2(2xy+3)}\\\\
y'+\frac{2y}{x}&=\frac{3x^2y^2+6xy+3-1}{x^2(2xy+3)}\\\\
y'+\frac{2y}{x}&=\frac{3(xy+1)^2-1}{x^2(2xy+3)}
\end{align*}\]
In either case, a substitution like \(v=yx\) would be nice.
\[v=yx~~\iff~~y=\frac{v}{x}~~\implies~~y'=\frac{xv'-v}{x^2}\]
Using the "simplified" equation,
\[\begin{align*}\frac{xv'-v}{x^2}&=\frac{2-v^2}{x^2(2v+3)}\\\\
xv'-v&=\frac{2-v^2}{2v+3}
\end{align*}\]
Hmm this looks like it'll work. You'll get a separable equation. I won't bother checking to see what the "complete-the-square" equation will yield, this way seems good enough.

Answer 10

\[\begin{align*}
xv'&=\frac{2-3v-3v^2}{2v+3}\\\\
\frac{2v+3}{2-3v-3v^2}~dv&=\frac{dx}{x}\\\\
\int\frac{2v+3}{\dfrac{11}{4}-3\left(v+\dfrac{1}{2}\right)^2}~dv&=\int\frac{dx}{x}\\\\
\int\frac{2\left(\sqrt{\dfrac{11}{12}}\sin u-\dfrac{1}{2}\right)+3}{\dfrac{11}{4}-3\left(\sqrt{\dfrac{11}{12}}\sin u\right)^2}~\left(\sqrt{\dfrac{11}{12}}\cos u\right)~du&=\int\frac{dx}{x}\\
&\color{gray}{\text{where }v+\frac{1}{2}=\sqrt{\frac{11}{12}}\sin u}\\\\
\int\frac{2\left(\sqrt{\dfrac{11}{12}}\sin u-\dfrac{1}{2}\right)+3}{\dfrac{11}{4}-3\left(\sqrt{\dfrac{11}{12}}\sin u\right)^2}~\left(\sqrt{\dfrac{11}{12}}\cos u\right)~du&=\int\frac{dx}{x}\\\\
\frac{\sqrt{\dfrac{11}{12}}}{\dfrac{11}{4}}\int\frac{2\sqrt{\dfrac{11}{12}}\sin u+2}{1-\sin^2u}~\cos u~du&=\int\frac{dx}{x}\\\\
\frac{\sqrt{\dfrac{11}{12}}}{\dfrac{11}{4}}\int\frac{2\sqrt{\dfrac{11}{12}}\sin u+2}{\cos u}~du&=\int\frac{dx}{x}\\\\
\frac{2}{3}\int\tan u~du+\dfrac{8}{\sqrt{132}}\int\sec u~du&=\int\frac{dx}{x}
\end{align*}\]
You get the idea...

Answer 11

Why does it look so complicated?

Answer 12

Probably due to the trig sub? The end result isn't too bad. Simplifying is a nightmare in LaTeX though.

Answer 13

I see. I'll work it out. Thank you for the help.

Answer 14

No problem! I'd encourage you to try your hand at a different approach (if there is one) for integrating if this seems too complicated.

Answer 15

But wait, I got xv'=(v^2+3v+2)/(2v+3). Why did you get the negative terms?

Answer 16

The smallest typos... >.<
\[\begin{align*}
y'&=\frac{2-x^2y^2}{x^2(2xy+3)}\\\\
\frac{xv'-v}{x^2}&=\frac{2-v^2}{x^2(2v+3)}\\\\
\color{red}{xv'}&=\color{red}{\frac{2-v^2}{2v+3}+v}\\\\
\color{red}{xv'}&=\color{red}{\frac{2+v^2+3v}{2v+3}}
\end{align*}\]
Sorry for the error. I guess I was used to the \(y=vx\) sub that gives \(y'=xv'+v\)...

Answer 17

Wait, so which substitution is right? v=yx or y=vx?

Answer 18

\[\Huge\text{We're using }{\bf\color{red}{v=yx}}\text{ here}.\]
In that case, there's a much simpler sub to use.
\[\begin{align*}
\int\frac{2v+3}{v^2+3v+2}~dv&=\int\frac{dx}{x}
\end{align*}\]
Set \(u=v^2+3v+2\), you get \(du=(2v+3)~dv\) and everything follows nicely from there.

Answer 19

Haha, thank you for clearing up the confusion. Now I got it.