0
$\begingroup$

The differential equation$$y′′′′+ay′′′+by′′+cy′+dy=0$$ has solution $$ y=−3te^{2t}+2e^{−2t}\sin(5t)$$

Find $a, b, c$ and $d$.


I've tried looking online for problems similar to this but had no luck. From what I imagine since there is only 2 and -2 in exponents, we can write $$(r - 2)(r + 2) = r^2 - 4$$

Does that mean $a = 0, b = 1, c = 0, d = -4$?

$\endgroup$
  • 2
    $\begingroup$ Start off with what it means for $y$ to be a "solution" to the differential equation. Compute the necessary number of derivatives of $y$ and write the left side of the equation in terms of exponentials, sines and cosines, with various coefficients. $\endgroup$ – user61527 Jun 8 '14 at 4:52
  • $\begingroup$ I acutally computed y(0) up until the 4th derivative, getting $$y(0) = 0, y'(0) = 7, y''(0)= -52, y'''(0) = -166, y''''(0) = 1584$$ $\endgroup$ – Freud Jun 8 '14 at 4:53
  • 2
    $\begingroup$ Looks like $r_1=-2+5i$ should be a zero of the characteristic polynomial, and $r_2=2$ should be a double zero. Therefore... $\endgroup$ – Jyrki Lahtonen Jun 8 '14 at 4:54
1
$\begingroup$

Knowing the solution $$y=−3te^{2t}+2e^{−2t}\sin(5t)$$ as suggested by user61527, you can just use brute force to get $y',y'',y''',y''''$ and plug their expressions into $$y′′′′+ay′′′+by′′+cy′+dy=0$$ After rearrangement, you should arrive to $$e^{-2 t} \left(-3 e^{4 t} (t (8 a+4 b+2 c+d+16)+12 a+4 b+c+32)+2 \sin (5 t) (142 a-21 b-2 c+d+41)+10 \cos (5 t) (-13 a-4 b+c+168)\right)=0$$ Cancelling all coefficients leads to the following equations $$32 + 12 a + 4 b + c=0$$ $$16 + 8 a + 4 b + 2 c + d=0$$ $$168 - 13 a - 4 b + c=0$$ $$41 + 142 a - 21 b - 2 c + d=0$$ for which the solutions are $a=0,b=17,c=-100,d=116$.

$\endgroup$
-1
$\begingroup$

Let D be the differential operator, that is, Dy = y'. The first term y1 in your equation for y satisfies (D-2)^2 y1 = 0. (The square takes care of the factor t.)

The second term y2 satisfies (D+2) y2 = 0.

The third term y3 satisfies (D^2+25) y3 = 0.

The sum y of all three terms satisfies

(D-2)^2 (D+2) (D^2+25) y = 0.

Multiply this out to get your answer.

Note: This is of order 5, not 4. I believe this is an error in the question. For example, if the second term were e^2t and not e^-2t then one would not need the D+2 factor.

$\endgroup$
  • 2
    $\begingroup$ $\exp(-2t)$ and $\sin(5t)$ are not solutions, $\exp(-2t) \sin(5t)$ is. That means you want a factor $(D+2-5i)(D+2+5i)$. $\endgroup$ – Robert Israel Jun 8 '14 at 5:48
  • $\begingroup$ Yes, this is correct. $$(r-2)^2(r+2−5i)(r+2+5i)$$ $\endgroup$ – Freud Jun 8 '14 at 6:00
  • $\begingroup$ Sorry. I seem to have inserted a "+" into the question. Anyway, the procedure is right. $\endgroup$ – berkeleychocolate Jun 8 '14 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.