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

MATH101 April 2005

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Question 01 (d)
Find the general solution $\displaystyle y=y(x)$ of the differential equation $\displaystyle y''-2y'+y=0.$

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!

Hint

This is a 2nd order differential equation with constant coefficients. So, you can begin by assuming that the general solution is the sum of two exponential functions, each of the form $y=ce^{rx}$, where $c$ is a constant. (Remember that there are special cases where solutions of the differential equation may look like sines and cosines, or may include a linear term in front of the exponential term.)

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. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This is a 2nd-order differential equation with constant coefficients. So make the substitution of $y=e^{rx}$ into the equation and solve for $r$. ${\begin{aligned}y''-2y'+y&=0\\r^{2}e^{rx}-2re^{rx}+e^{rx}&=0\\e^{rx}\left(r^{2}-2r+1\right)&=0\\\rightarrow \quad e^{rx}=0\quad {\mbox{or}}&\quad r^{2}-2r+1=0\end{aligned}}$ Solving for $r$ we find that there is a double root $r=1$. Thus we have found that one solution is $y(x)=e^{x}$, but we need to find a second solution. We multiply the first solution by $x$ and confirm that it is indeed a solution of the above equation (i.e: Let $y=xe^{x}$ be another possible solution and substitute it into the differential equation). ${\begin{aligned}y''-2y'+y&=\left(2e^{x}+xe^{x}\right)-2\left(xe^{x}+e^{x}\right)+xe^{x}\\&=2e^{x}+xe^{x}-2xe^{x}-2e^{x}+xe^{x}\\&=0\end{aligned}}$ Thus, the general solution is $\,y(x)=c_{1}e^{x}+c_{2}xe^{x}$, where $c_{1},\,c_{2}$ are arbitrary constants.

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