Could use a little advice here, with a 2nd order ODE

[MaximumTaco]

xy'' -(2x+1)y' + (x+1)y = (x ex)2

I know a solution - (x-1)e2x

Thus, y= ((x-1)e2x u(x))

Now, i know how to do the whole reduction of order thing, but when i find y' and y'' and substitute, the u(x) term doesn't cancel out so this doesn't work

(x2-x)u'' + (2x2-x+1)u' + x2u = x2

So, how do i approach this? Thanks.

[ReyChiquito]

You know that the solution can be written in the form y(x)=y_{p}(x)+y_{h}(x) where y_{p}(x) is the particular solution and y_{h}(x) is the solution to the homogeneous equation

xy_{h}''(x)-(2x+1)y_{h}'(x)+(x+1)y_{h}(x)=0.

You already know that y_{p}(x)=(x-1)e^{2x}.

For the homogeneous equation, there are two linearly independent solutions, one of them is clearly

y_{h_{1}}(x)=c_{1}e^{x}

and using the reduction of order, the other is

y_{h_{2}}(x)=c_{2}\frac{x^2}{2}e^{x}.

So the general solution to the ode is
y(x)=c_{1}e^{x}+c_{2}\frac{x^2}{2}e^{x}+(x-1)e^{2x}

If you need further justification, use frobenius method for the homogenous equation to calculate y_{h_{1}}(x), reduction of order for y_{h_{2}}(x), and variation of parameters for y_{p}(x). That way the problem is fully justified.

ps. i deleted my first post as it was wrong and confusing.

[MaximumTaco]

I understand the first technique, but i'm not familiar with the Frobenius method, i looked it up on Mathworld and it looks really complex

With your other post, i can't quite follow how to get the second linearly independant solution. Thanks.

[ReyChiquito]

Ignore the first post... is nonsence, thats why i deleted it.

Given y_{1}(x)=e^{x} using reduction of order to get the second solution

y_{2}(x)=v(x)e^{x}

implies that

y'_{2}(x)=[v'(x)+v(x)]e^{x}

y''_{2}(x)=[v''(x)+2v'(x)+v(x)]e^{x}

substituting in xy''_{2}(x)-(2x+1)y'_{2}(x)+(x+1)y_{2}(x)=0

we get that

xv''(x)-v'(x)=0

do i need to go further?

ps. Frobenius method is used only if you want to *construct* the first solution y_{1}=c_{1}e^{x}. Just so you know that it didnt came from divine inspiration, you dont need to use it though.