Finding the coefficient of an ODE

Question

The real constants $_3,_2,_1,_0$ are such that the ODE $$^{(4)}+_3^{(3)}+_2''+_1'+_0=0$$ has $3^{−}+^{−4}\sin()$ as a solution. What is $_0$?

I would really like some help in solving this problem.

The hint was to figure out the characteristic polynomial first, which is $$r^4 + a_3r^3 + a_2r^2 + a_1r + a_0 = 0.$$ While I could plug in the given solution in, that'd be too complicated. I've tried to see if I could simplify the expression $3^{−}+^{−4}\sin()$ further into one term, but so far haven't really found a way to do so. Any help would really be appreciated!

Answer 1

The $te^{-t}$ means that $-1$ is a repeated root. The $e^{-4t}\sin t$ means that $-4+i$ and its conjugate are roots because complex roots always come in conjugate pairs for real polynomials. This is a degree $4$ polynomial hence we know we are done looking for roots.

The $a_0$ term of the polynomial is the product of all of its roots times the coefficient of the leading term (imagine what the last constant term would be if we multiplied out $(r-r_1)\cdots (r-r_n)$ for an even degree polynomial we have factored by its roots while the leading coefficient is $1$) so we have that

$$a_0 = 1\cdot(-1)^2\cdot(-4+i)\cdot(-4-i) = 17$$

We can also find by pattern recognition that $a_3$ is the negative sum of the roots, $a_2$ the positive sum of the product of any two roots, and $a_1$ the negative sum of the product of any three roots, giving us

$$a_3 = -(-2-8) = 10$$

$$a_2 = 1 + 17 + 2(4-i) + 2(4+i) = 34$$

$$a_1 = -(-8-34) = 42$$

Answer 2

Let $P(r)$ be the characteristic polynomial mentioned in the OP. Because of the form of the special solution, we know that $-1$ is a double solution of $P$ (this is because of the $t\exp(-t)$, and the other term, a linear combination of $\exp((-4\pm i)t)$, shows that $-4\pm i$ are also roots. So $P$ is $$ P(r)=(r-(-1))^2(r-(-4+i))(r-(-4-i))\ . $$ The free coefficient is $(-1)^2(-4+i)(-4-i)=17$.

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Computer check, here using sage:

sage: var('r');
sage: P = (r-(-1))^2 * (r-(-4+i)) * (r-(-4-i))
sage: P.expand()
r^4 + 10*r^3 + 34*r^2 + 42*r + 17
sage: var('t');
sage: f = t*exp(-t) + exp(-4*t)*sin(t)
sage: E = diff(f,t,4) + 10*diff(f,t,3) + 34*diff(f,t,2) + 42*diff(f,t) + 17*f
sage: E.simplify_full()
0

Answer 3

One wonders whether if it is a good idea to first solve the ODE. It is straightforward to put the known solution into the ODE and to identify the coefficients.

From $y=3te^{-t}+e^{-4t}\sin(t)$ compute $y^{(1)}, y^{(2)}, y^{(3)}$

Put them into the ODE and identify the coefficients of the different functions of $t$ which must equal to $0$ so that the sum of the terms equal $0$ any $t$. This leads to the equations : $$-3a_3+3a_2-3a_1+3a_0+3=0$$ $$9a_3-6a_2+3a_1-12=0$$ $$47a_3-8a_2+a_1-240=0$$ $$-52a_3+15a_2-4a_1+a_0+161=0$$ After simplification and solving the linear system: $$a_3=10\quad;\quad a_2=34\quad;\quad a_1=42\quad;\quad a_0=17.$$