Question

I just need a nudge in the right direction on this differential equation...

Answer 1

ok

Answer 2

The equation I need to solve is:

\[2x^3y'=1+\sqrt{1+4x^2y}\]

I tried integration factor but the integrals are really nasty.

Answer 3

I just need advice on how to approach it. I don't want a solution.

Answer 4

we can solve for y

Answer 5

\[2x^3y'=1+\sqrt{1+4x^2\color{red}{y}}\]
or
\[2x^3y'=1+\sqrt{1+4x^2}~\color{red}{y}~~?\]
The first equation is not linear, so you can't find an integrating factor (unless you can make an appropriate substitution).

Answer 6

The first case. And you can find an integrating factor, it is just not of the usual form.

Taking my equation I can write it in the form:

\[A(x,y)dx+B(x,y)dy=0\]

Which is all that is needed for an integrating factor and I solved for it. But then using the integrating factor to solve the now exact equation is extremely tough.

Answer 7

Oh sorry I assumed you meant linear integrating factor...

Answer 8

So you've done this so far, correct?
\[2x^3y'=1+\sqrt{1+4x^2y}~~\iff~~\left(1+\sqrt{1+4x^2y}\right)~dx-2x^3~dy=0\]

Answer 9

For reference, my integrating factor comes out to be:

\[\frac{1+\sqrt{1+4x^2y}}{2\sqrt{y}x^4}\]

which I have checked in Mathematica. Now I need to solve:
\[\frac{\partial \Phi}{\partial x} = \lambda(x) A(x,y); \frac{\partial \Phi}{\partial y} = \lambda(x)B(x,y)\]
Find Phi is very difficult with the given functions.

Answer 10

Without resorting to Mathematica to do my integrals***

Answer 11

Also @satellite73

Answer 12

http://www.wolframalpha.com/input/?i=solve+2x^3*y%27+%3D+1+%2B+sqrt%28+1+%2B+4x^2*y%29

Answer 13

The integrating factor should be one-dimensional for the equation to becomes exact (in terms of \(x\) or \(y\) only).

Answer 14

you can make a substitution