Science:Math Exam Resources/Courses/MATH101/April 2005/Question 01 (e)/Solution 1

To find the general solution for this problem, we need to find both the particular solution, $y_{p}$, and homogeneous solution, $y_{h}$.

The homogeneous solution satisfies

$\displaystyle {}y_{h}''-2y_{h}'+y_{h}=0.$

From part (d) of this exam we have its general solution

$\displaystyle y_{h}(x)=c_{1}e^{x}+c_{2}xe^{x}.$

To find the particular solution, we look at the right hand side of the equation and notice that there is a term that is linear in $x$ (as opposed to quadratic, cubic, etc.). So we can assume a trial solution of $y_{p}=ax+b$ and plug this into the differential equation to find the constants $a,b$ such that the original differential equation is satisfied.

${\begin{aligned}y_{p}''-2y_{p}'+y_{p}&=x\\0-2a+ax+b&=x\end{aligned}}$

By comparing the linear and constant terms on both sides of the equation, we can see that for the differential equation to be satisfied, we must satisfy $a=1$ and $b=2a=2$.

So the general solution to the differential equation is the sum of the homogeneous and particular solutions:

$\displaystyle y(x)=x+2+c_{1}e^{x}+c_{2}xe^{x}$

where $c_{1},\,c_{2}$ are arbitrary constants.