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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $a=0,b=17,c=-100,d=116$.

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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.

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    $\begingroup$ $\exp(-2t)$ 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

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