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

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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.
 
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I don't think you calculated (x^2e^{ix})'' correctly. Please show your working for this.
 
pasmith said:
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:
$$
\begin{align*}
A' x^2 e^{ix} + 2B' x e^{ix} + \iota ~ B' x^2 e^{ix} + 2C' e^{ix} + \iota~ 4C' x e^{ix} - C' x^2 e^{ix} \\
(A'+ \iota~ B' -C') x^2 e^{ix} + (2B' + \iota 4C') x e^{ix} + 2C' e^{ix} \\
A x^2 e^{ix} + B x e^{ix} + C e^{ix} \\
(Ax^2 + Bx + C)e^{ix} \\
\end{align*}
$$
 
If y_p = Ax^2 e^{ix} + Bxe^{ix} then <br /> y&#039;&#039;_p + y_p = A(2 + 4ix)e^{ix} + B(2i)e^{ix}. Setting that equal to 4xe^{ix} gives <br /> \begin{split}<br /> 2A + 2iB &amp;= 0 \\<br /> 4iA &amp;= 4.\end{split}
 
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