User:Egm6321.f2010.team5.riveros/hw5p4

=Problem 4=

Given
We are given the differential equation,
 * $$\displaystyle y'+P(x)y=Q(x)$$

where the homogeneous solution has been previously found to be,
 * $$\displaystyle y_h=Ae^{-\int{P(x)dx}}$$

Find
We are to find a particular solution, $$y_p$$, by variation of parameters.

Solution
We first let
 * $$\displaystyle y(x)=u(x)\cdot y_h$$

By taking the derivative we find,
 * $$\displaystyle y'(x)=e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]$$

We substitution this into the given equation,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]+P(x)u(x)e^{-\int{P(x)dx}}=Q(x)$$

Rearranging yields,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)+u(x)P(x)]=Q(x)$$

Which then simplifies to,
 * $$\displaystyle e^{-\int{P(x)dx}}u'(x)=Q(x) $$

We solve for $$u'(x)$$
 * $$\displaystyle u'(x)=y_{h}^{-1}\cdot Q(x)=\exp [\int{P(x)dx}]\cdot Q(x) $$

Integrating,
 * $$\displaystyle u(x)=\int{\exp [\int{P(x)dx}]\cdot Q(x)dx} $$

Therefore, the particular solution is
 * $$\displaystyle y(x)=exp [-\int{P(x)dx}]\cdot \int{\exp [\int{P(x)dx}]\cdot Q(x)dx}$$