general solution of differential equation

[b]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.

[b]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?

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 use integration by parts
 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?

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general solution of differential equation

 Quote by captainjack2000 [b]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. [b]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?
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)!

 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