0
$\begingroup$

Given that $y = 3e^{-2x}$ and $y = 8e^{-4x}$ are solutions of the differential equation $$ay'' + 2by' + cy = 0$$
where a, b, c are non-zero constants. Which one of the following equations must hold?

A. $2b + c = 0$
B. $3a - b = 0$
C. $6a - c = 0$
D. $8b + 3c = 0$

I have re-written the differential equation as $$y'' + \frac{2b}{a}y' + \frac{c}{a}y = 0$$ and also found out that given the two solutions, I can find the corresponding factorized auxiliary equation $$ (\lambda + 2)(\lambda + 4) = 0$$
may I know how I should proceed to compare the two equations as I can't seem to find a basis to match constants.

  • 0
    $\dfrac{2b}{a}=6$ and $\dfrac{c}{a}=8$. – 2017-02-22
  • 0
    @MyGlasses Hi again, thanks for your constant help! Did you by any chance get to there by matching $\frac{2b}{a}$ as the negative sum of roots and $\frac{c}{a}$ as the product of roots? – 2017-02-22
  • 0
    @MyGlasses Ah.. I'm curious how did you get $\frac{2b}{a} = 6$? – 2017-02-22
  • 0
    From the factorization you get that $a:2b:c = 1:6:8$, all other proportionalities derive from this combined one. – 2017-02-22
  • 0
    @LutzL May I know how you derived at that proportionality? Sorry there must be something that I haven't learnt or put to use adequately to understand... – 2017-02-22
  • 0
    Well, your answer is $y'' + \dfrac{2b}{a}y' + \dfrac{c}{a}y = 0$ so characteristic equation is $\lambda^2 + \dfrac{2b}{a}\lambda + \dfrac{c}{a} = 0$ and your factorized it by $ (\lambda + 2)(\lambda + 4) =\lambda ^2+6\lambda +8= 0$ compare them. – 2017-02-22
  • 0
    Via the factorization you found that the given functions are solutions of $y''+6y'+8y=0$. As the orders match, the differential equations can only differ by a factor. Thus $(a,2b,c)=\lambda·(1,6,8)$ which immediately gives $\lambda=a$,... – 2017-02-22
  • 0
    @MyGlasses My bad, how could I have forgotten that I can express the original D.E. in the characteristic equation format for comparison... Silly and careless me. Thank you for pointing that out! – 2017-02-22
  • 0
    Your welcome... – 2017-02-22
  • 0
    @LutzL Thank you! I've never really tried to look at equations as a factor of one another in the way you mentioned, you gave me a new perspective to look at things. – 2017-02-22

1 Answers 1

3

By comparing the characteristic equation with the roots, we can find. $$\frac{2b}{a} = 6$$ $$\frac{c}{a} = 8$$
Therefore $2b = 6a$, which gives us $b = 3a$.
With this identity we can then formulate that $3a - b = 0$.

Answer is B.