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

MATH101 April 2005

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) •

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[hide] Question 01 (e)
Find the general solution $\displaystyle y=y(x)$ of the differential equation $y^{\prime \prime }-2y^{\prime }+y=x$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
To find the general solution for this NON-HOMOGENEOUS differential equation, you must find both the homogeneous and particular solution. You can find the particular solution by assuming a trial solution inspired by the right hand side of the differential equation.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
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[show] Solution
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.