# Thread: Little help: Particular Integral for this equation

1. ## Little help: Particular Integral for this equation

Im solving:

y''-2y'+y= (2e^x)(cosx)

The complimentary function after solving the auxiliary equation is:

(A+Bx)e^x

How should I start finding out the particular integral?

by substituting

y= k(e^x)(cosx) ( where only k is a constant)

or y= k(e^p.x)(cos q.x) ( where k, p and q are constants)

2. Are you familiar with Variation of Parameters?.

$\displaystyle y''-2y'+y=2e^{x}cos(x)$

$\displaystyle m^{2}-2m+1=0\Rightarrow m=1$

Because 1 has multiplicity 2, we have:

$\displaystyle y_{c}=C_{1}e^{x}+C_{2}xe^{x}$

Note that $\displaystyle y_{1}=e^{x}, \;\ y_{2}=xe^{x}$

Wronskians:

$\displaystyle W=\begin{vmatrix}e^{x}&xe^{x}\\e^{x}&(x+1)e^{x}\en d{vmatrix}=e^{2x}$

$\displaystyle W_{1}=\begin{vmatrix}0&xe^{x}\\2e^{x}cos(x)&(x+1)e ^{x}\end{vmatrix}=-2xcos(x)e^{2x}$

$\displaystyle W_{2}=\begin{vmatrix}e^{x}&0\\e^{x}&2e^{x}cos(x)\e nd{vmatrix}=2e^{x}cos(x)$

$\displaystyle {u'}_{1}=\frac{W_{1}}{W}=-2xcos(x)$

$\displaystyle {u'}_{2}=\frac{W}{W_{2}}=2cos(x)$

Integrate the last two and get:

$\displaystyle u_{1}=-2(cos(x)+xsin(x))$

$\displaystyle u_{2}=2sin(x)$

$\displaystyle u_{1}+u_{2}=-2e^{x}cos(x)$

So, put them together with $\displaystyle y_{c}$ and get

$\displaystyle \boxed{y=C_{1}e^{x}+C_{2}xe^{x}-2e^{x}cos(x)}$

3. Can you point me to some notes/slide on the Wronskians method.

4. Do you have a DE book?. It will be in there under Variation of Parameters. Just google it. It's easy. It's just derivatives and determinants.