General solution of differential equation

1. Oct 16, 2007

captainjack2000

1. The question asks me to show that e^x is a solution of xy'' - (2x+1)y' + (x+1)y=0 and find the general solution.

2. I managed to simplify the equation to u''xe^(x) - u'e^(x) = 0 by letting y=ue^(x) and finding the differentials and substituting them in.
I've then let z=u dz/du=u' and d^2z/du^2 = u''
so I get xe^(x)(d^2z/du^2) - e^(x)dz/du = 0
How would I solve this?

2. Oct 16, 2007

kring_c14

use integration by parts

3. Oct 16, 2007

captainjack2000

Sorry but I am still confused;

if xe^(x)(d^2z/du^2) - e^(x)dz/du = 0
can I simplify this to

x(d^2z/du^2) = dz/du

xdz = du
xz = u +c but z = u???
xu = u +c
How does this help find the general solution?

4. Oct 17, 2007

HallsofIvy

Staff Emeritus

You don't solve that! It has two independent variables, u and x. Surely, you don't mean "let z=u dz/du=u' and d^2z/du^2 = u''. If you let z= u, then dz/du= 1. Perhaps you meant "let z= u, dz/dx= u'". But in that case you've gained nothing- you've just renamed u. Much better is "let z= du/dx, dz/dx= d2u/dx2". Then your equation becomes the first order equation xe^(x) z'- e^(x) z= 0. I would be inclined to first divide the entire equation by e^(x)!

5. Apr 30, 2011

doroulla

Hi can someone please help me with: general solution of dy/dx - y = x + 2x^2

i know how to find general solutions but only when i can seperate the y and x in 2 sides and multiply with dx and dy. someone please help me. i have looked in many books

6. Apr 30, 2011

HallsofIvy

Staff Emeritus