Originally Posted by

**Altair** I have to find the particular solution of the following DE.

$\displaystyle D^2 - 3D + 2 = sinx$

Here is how I did it.

$\displaystyle \frac{sinx}{D^2-3D+2}$

$\displaystyle \frac{Im(e^{ix})}{D^2-3D+2}$

$\displaystyle \frac{cosx + isinx}{(i)^2 -3(i) +2}$

$\displaystyle \frac{cosx + isinx}{1-3i}$

Then after rationalizing,

$\displaystyle \frac{cosx-3sinx}{10}$

Where as the answer given by the book is,

$\displaystyle \frac{sinx + 3cosx}{10}$

Can you please tell where is my mistake ?