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

[MaximumTaco]

xy'' -(2x+1)y' + (x+1)y = (x e^x)²

I know a solution - (x-1)e^2x

Thus, y= ((x-1)e^2x 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 independent solution. Thanks.

 

[ReyChiquito]

Ignore the first post... is nonsence, that's 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 don't need to use it though.