General Solution to Linear Homogenious Differential Equation

[nivekious]

I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?

 

[galactus]

Re: General Solution to Linear Homogenious Differential Equa

Is that the original problem?. If not, please post the problem the way it was given.

If this is what you have:

\(\displaystyle e^{-x^{2}}u''=u+xu'\)

then, we could write

\(\displaystyle u''=e^{x^{2}}(u'x+u)\)

This apears to have been a substitution DE.

an example:

\(\displaystyle (x^{2}+y^{2})dx+(x^{2}-xy)dy=0\)

Let \(\displaystyle y=ux, \;\ dy=udx+xdu\)

\(\displaystyle (x^{2}+u^{2}x^{2})dx+(x^{2}-ux^{2})[udx+xdu]=0\)

\(\displaystyle x^{2}(1+u)dx+x^{3}(1-u)du=0\)

\(\displaystyle \frac{1-u}{1+u}du+\frac{1}{x}dx=0\)

After integrating and using log properties, we get:

\(\displaystyle (x+y)^{2}=cxe^{\frac{y}{x}}\)

Is this the sort of thing you're dealing with?.