Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 20

MATH100 A December 2023

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[hide] Question 20
Find all values of ${\textstyle r}$ such that ${\textstyle y=e^{rt}}$ solves the differential equation $$y''+6y'+5y=0.$$

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!

[show] Hint
Since the function $y(t)=e^{rt}$ solves the differential equation, this means that when we substitute ${\textstyle y=e^{rt}}$ into the left-hand side of the differential equation, the result is the constant 0 function. After substituting, can you write the left hand side of the equation as a product of a polynomial and a function that is always positive?

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[show] 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. Let’s start by substituting ${\textstyle y=e^{rt}}$ into the differential equation. To do this, first calculate ${\textstyle y'=re^{rt}}$ and ${\textstyle y''=r^{2}e^{rt}}$. Then $${\begin{aligned}y''+6y'+5y=r^{2}e^{rt}+6re^{rt}+5e^{rt}&=0\\e^{rt}\left(r^{2}+6r+5\right)&=0\end{aligned}}$$ Since ${\textstyle e^{rt}}$ is always positive, we can divide through by it to get: $$r^{2}+6r+5=0,$$ which is a simple quadratic equation! It factors as ${\textstyle r^{2}+6r+5=(r+5)(r+1)=0}$, so the values of ${\textstyle r}$ such that ${\textstyle y=e^{rt}}$ solves the differential equation are ${\textstyle r=-1,-5}$.