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

y_{h^{″}} - 2 y_{h^{'}} + y_{h} = 0 .

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

y_{h} (x) = c_{1} e^{x} + c_{2} x e^{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 $y_{p} = a x + b$ 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^{″}} - 2 y_{p^{'}} + y_{p} & = x \\ 0 - 2 a + a x + 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 $b = 2 a = 2$ .

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

y (x) = x + 2 + c_{1} e^{x} + c_{2} x e^{x}

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