- #1

Hall

- 351

- 88

- 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.

$$

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.