Originally Posted by

**AllanCuz** There should be a standard table in your text that outlines what you should guess given a particular L.H.S

In this case,

$\displaystyle L.H.S = xe^{-x/2} + xcos(x/2) $

But let us note that we can use the prinipal of superposition here so

$\displaystyle L.H.S_1 = xe^{-x/2}$

$\displaystyle L.H.S_2 = xcos( \frac{x}{2} ) $

Finally,

$\displaystyle L.H.S= L.H.S_1 + L.H.S_2 $

So lets find an appropriate $\displaystyle Y_P$ for our L.H.S

For $\displaystyle L.H.S_1$ we note that we have the form

$\displaystyle f(x)e^{g(x)}$

Which by standard tables suggest we try the form

$\displaystyle Y_{P1} = (Ax+B)e^{g(x)} $

For $\displaystyle L.H.S_2$ we note that we have the form

$\displaystyle f(x)cos( g(x) ) $

Which, again by stanard tables suggests we try the form

$\displaystyle Y_{P2} = (Cx + D)cos( p(x) ) + (Ex+F) sin ( p(x) ) $

We include both Sin and Cos here because both are potential solutions (remember our trig circles!). Everytime you have sin or cos as your left hand side, you guess the appropriate function for both sin and cos.

In the end,

$\displaystyle Y_P= Y_{P1} + Y_{P2} = (Ax+B)e^{g(x)} (Cx + D)cos( p(x) ) + (Ex+F) cos ( p(x) ) $

$\displaystyle (Ax+B)e^{ \frac{-x}{2} } (Cx + D)cos( \frac{x}{2} ) + (Ex+F) cos ( \frac{x}{2} ) $

EDIT....I went through this entire thing use L.H.S when in fact we are using the R.H.S but it is late so dont castrate me lol