Mr. Tenenbaum's and Prof. Mattuck's advice not working (ODE)

[Hall]

Homework Statement

$$
y'' + y = 4x \sin x
$$

Relevant Equations

Case 2: Q(x) contains a term which, ignoring constant coefficients, is $x^k$ times a term $u(x)$ of $y_c$, where $k$ is zero or a positive integer. In this case a particular solution $y_p$ of $(21.1)$ will be a linear combination of $x^{k+1} u(x)$ and all its linearly independent derivatives.

All right, we got
$$
y'' + y = 4x \sin x
$$

We are doing the Complexification
$$
\tilde{y''} + \tilde{y} = 4x e^{ix}
$$
Complementary function:
$$
\begin{align*}
\textrm{characteristic equation =}\\
m^2 + 1 = 0 \\
m = \pm i \\
\tilde{y_c} = c_1 e^{ix} + c_2 e^{-ix} \\
\end{align*}
$$

Q(x), that is RHS of the original non-homogeneous equation, is xeix (ignoring constant coefficients), therefore,
$$
\begin{align*}
\tilde{y_p} = (Ax^2 + Bx + C) e^{ix} \\
\tilde{y_p }'' + \tilde{y_ p} = 2 A e^{ix} + 2 i (Ax+B) e^{ix} \\
4x = 2A + 2iAx + 2B \\
\implies A = -2 i , ~B = 2 i \\
\end{align*}
$$

Hence,
$$
\begin{align*}
\tilde{y_p} = (-2 i x^2 + 2i x + C) e^{ix} \\
\textrm{We can leave that C because it is already there in}~y_c \\
Im(\tilde{y_p})= y_p = -2 x^2 \cos x + 2x \cos x \\
\end{align*}
$$

The answer that I have got is wrong, but I don't know where the mistake lies.

 

[pasmith]

I don't think you calculated $(x^2e^{ix})''$ correctly. Please show your working for this.

 

[Hall]

I don't think you calculated $(x^2e^{ix})''$ correctly. Please show your working for this.

$$
\begin{align*}
(x^2 e^{ix})' = 2xe^{ix} + \iota x^2 e^{ix} \\
(x^2 e^{ix})'' = 2 e^{ix} + \iota ~ 2x e^{ix} + \iota 2 x e^{ix} - x^2e^{ix} \\
(x^2 e^{ix})'' = 2 e^{ix} + \iota~ 4x e^{ix} - x^2e^{ix}
\end{align*}
$$
A linear combination of $x^2 e^{ix}$, $2x e^{ix} + \iota x^2 e^{ix}$, and $2 e^{ix} + \iota~4x e^{ix} -x^2e^{ix}$ would give us something like this:
$$
\begin{align*}
A' x^2 e^{ix} + 2B' x e^{ix} + \iota ~ B' x^2 e^{ix} + 2C' e^{ix} + \iota~ 4C' x e^{ix} - C' x^2 e^{ix} \\
(A'+ \iota~ B' -C') x^2 e^{ix} + (2B' + \iota 4C') x e^{ix} + 2C' e^{ix} \\
A x^2 e^{ix} + B x e^{ix} + C e^{ix} \\
(Ax^2 + Bx + C)e^{ix} \\
\end{align*}
$$

 

[pasmith]

If $y_p = Ax^2 e^{ix} + Bxe^{ix}$ then $$y''_p + y_p = A(2 + 4ix)e^{ix} + B(2i)e^{ix}.$$ Setting that equal to $4xe^{ix}$ gives $$\begin{split} 2A + 2iB &= 0 \\ 4iA &= 4.\end{split}$$