Originally Posted by

**topsquark** $\displaystyle y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

In line 3, that should be "-10 - 20i," not "6 - 36i."

In line 4 you need to rationalize the denominator, not just drop the real part. Once you have rationalized, then use [tex]e^{4ix} = cos(4x) + i \, sin(4x), then expand the product, THEN take the imaginary part.

So line 4 should read:

$\displaystyle y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = 2 \mathbb{IM} \left (e^{4ix} \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right ) \right )$

= $\displaystyle 2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

= $\displaystyle 2 \mathbb{IM} \left ( -cos(4x) \cdot \frac{1}{50} - i \, sin(4x) \cdot \frac{1}{50} + cos(4x) \cdot \frac{1}{25}i + i \, sin(4x) \cdot \frac{1}{25}i \right )$

= $\displaystyle 2 \left ( \frac{1}{25}cos(4x) - \frac{1}{50}sin(4x) \right )$

= $\displaystyle \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

-Dan