Question

solve : 1) (x+y^(2) * x) dx + e^(x^e) y dy=0

2) y-(2x-4y^2)dy/dx=0

Answer 1

\[(x+y^2x)dx+(e^{x^e}y)dy=0\] is the first?

Answer 2

yes

Answer 3

Separable... no. Integrating factor... not any obvious ones... euler equations... no. Gauss's equations.. no. My goodness. I honestly don't know, this looks to be something beyond a first year ordinary linear differential equation.

What class is this for?

Answer 4

Yeah I got nothing...

Answer 5

They're both nonlinear. I guess you can do numerical approximations on them if you have a decent idea of what might be a solution. I can't think of any physical situations where such differential equations might arise. I'm curious to know what class these equations come from.

http://math.exeter.edu/rparris/winplot.html is a great freeware program which should let you qualitatively analyze them (graph them, see specific solutions, etc.). It also has a good numerical solver.

Sorry I can't be much help beyond that.

Answer 6

The first equation is seriously ugly and doesn't have a solution in terms of elementary functions; although you can force it look like this

\[2yy' - 2xe^{x^e}y^2 = -2xe^{x^e}\]
The left hand side can now be made exact by multiplying through by
\[H(x) = \exp \left[ \int\limits -2xe^{x^e} dx \right]\]
so we have
\[\left(y^2 H(x) ) \right)' = -2xe^{x^e}H(x)\]
and you can take it from there.

With a little manipulation you can make H(x) look like something involving the Gamma Function, but it is well known that function is not reducible to elementary functions.

Answer 7

The second equation is slightly easier but I'm also convinced it doesn't have a solution in terms of elementary functions.

I thought of this second equation first as an equation in x = x(y) and dx/dy. Then it became relatively easy to solve for x. Inverting that resulting equation x=x(y) is unfortunately not straight-forward.

Answer 8

Dang James you are the man. Wow...

Answer 9

thanks , i will see wat i will get but as i got this equation need to do this equation through linear due to y or x or Bernoulli or Separable one of these ways :S

Answer 10

Respect to James. :) Nicely done!

Answer 11

To confirm also, the second equation is
\[{dy \over dx} = {y \over 2x-4y^2}\]
yes?

Answer 12

yes

Answer 13

I was hoping it would be a Bernoulli equation for a moment, but you can there's no obvious transformation to put this in Bernoulli form.

Answer 14

then i guess we make it dx/dy = (2x-4y^2 )/ y

Answer 15

Right, that's what I did and now it's straightforward

Answer 16

yes :)