What is the particular integral for y''+y'+y=xe^{2x}?

[Air]

When finding the particular integral, I understand that for:

$$y''+y'+y=e^{2x}$$, One would choose $$y=Ae^{2x}$$
$$y''+y'+y=x$$, One would choose $$y=Ax+B$$

But what am I supposed to choose if I have: $$y''+y'+y=xe^{2x}$$?

 

[Count Iblis]

When finding the particular integral, I understand that for:

$$y''+y'+y=e^{2x}$$, One would choose $$y=Ae^{2x}$$
$$y''+y'+y=x$$, One would choose $$y=Ax+B$$

But what am I supposed to choose if I have: $$y''+y'+y=xe^{2x}$$?

Well, you know that in case of:

Dy = exp(p x)

where D is a linear differential operator,

you would choose y = A exp(px)

You also know that a solution of:

Dy = f(x) + g(x)

can be obtained by solving

Dy = f(x)

and

Dy =g(x)

separately and adding up the solutions. Linearitity of D implies that this will work.

Then, in case of:

Dy = x exp(px)

you could simply consider the factor x to be the derivative w.r.t. p. So, if you simply solve:

Dy = exp(px)

by putting y = A(p)exp(px), then the linearity of D implies that the solution of

Dy = [exp[(p+epsilon)x] - exp(px)]/epsilon

is given by

[A(p+epsilon)exp[(p+epsilon)x] - A(p)exp(px)]/epsilon

So, taking the derivative of A(p)exp(px) w.r.t. p will do. By Leibnitz's rule, you see that this amounts to putting

y = (A + B x)exp(px)

 

[Random Variable]

[itex] (A + Bx) e ^{2x} [/tex]

for [itex] x^{2} e^{2x}[/tex] use $(A + Bx +Cx^{2})e^{2x}$

and so on