Question

Find the equations of the trajectories of y"+y^3=0.

Answer 1

@SithsAndGiggles

Answer 2

Do you know how to draw up the phase portrait?

Answer 3

No.

Answer 4

The answer is v^2+y^4/4=C
I guess that p(y)=y^3,
integrate p(y) to obtain P(y),
P(y)=y^4/4
But I don't know how to get v^2.

Answer 5

y"+p(y)=0
v(dv/dy)+p(y)=0
integrate
v^2/2+P(y)=C

Answer 6

I can't say I'm at all familiar with this method :/

Answer 7

Yeah, the answer is pretty weird. If following the formula above, the answer should be v^2/2+y^4/4=C, but the answer in the book says v^2+y^4/4=C.

Answer 8

So are we assuming \(v\) is the derivative of \(y\)?

Answer 9

Well, I guess so. According to the textbook, it says y"=v(dv/dy).

Answer 10

Or is \(v\) a function of \(y\)?

Answer 11

Assuming \(v=v(y)\), differentiating the given solution with respect to \(y\), you have
\[\frac{d}{dy}\left[v^2+\frac{y^4}{4}\right]=0~~\implies~~2v\frac{dv}{dy}+y^3=0\]
which would suggest that \(y''=2v\dfrac{dv}{dy}\).

Answer 12

So from y"+y^3=0, how would you do this problem?

Answer 13

One thing I might try would be to multiply both sides by the first derivative:
\[y'y''+y^3y'=0\]
Then notice that reverse chain rule gives
\[\begin{align*}\frac{d}{dt}\left[\frac{(y')^2}{2}\right]+\frac{d}{dt}\left[\frac{y^4}{4}\right]&=0\\\\
\frac{1}{2}(y')^2+\frac{1}{4}y^4&=C\end{align*}\]
But past this point, I'm not sure how this could be solved...

Answer 14

Quote
`Yeah, the answer is pretty weird. If following the formula above, the answer should be v^2/2+y^4/4=C, but the answer in the book says v^2+y^4/4=C`

Your book has a small typo, it should be v^2/2 + y^4/4 = C

This will give you:
(dy/dx)^2/2 + y^4/4 = C

See siths and giggles last tell how to get that answer using reverse chain rule . Books sometimes make typos :)

Answer 15

Thank you @SithsAndGiggles and @perl !

Answer 16

solving it further is actually pretty difficult. it can be a fun challenge problem. First solve for dy/dx

(dy/dx)^2/2 + y^4/4 = C

(dy/dx)^2/2 = C - y^4/4

(dy/dx)^2/2 = 2C - y^4/2

dy/dx = + / sqrt (2C - y^4/2

dy / sqrt( 2C - y^4/2) = +/- dx

You can integrate both sides, and you get what is known as Jacobi Elliptical integral