Higher Order Differential Equation Solution 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 $a = 0, b = 1, c = 0, d = -4$?
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 $a=0,b=17,c=-100,d=116$.
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.