Higher Order ODE with Differential Operators

Question

I am trying to solve an ODE problem involving higher order.

Let $p(s) = s(s^2-s+1)(s-1)$ and $D = d/dt$. Solve the initial value problem $$p(D)x = t + e^t,$$ $x'''(2) = 1$, $x''(2) = 1$, and $x'(2) = 1$, and $x(2) = 0$.

Attempt: I believe I need to solve the homogenous equation by finding all the roots of $p(s)$. Then, for the particular solution I don't really know what to do? Variation of parameters? (Method of Undetermined Coefficient maybe?, I dislike that method and would like to avoid it) .

Edit: Also what is $p(s)$, and $p(D)x$, I am confused on those two things as-well.

Jessica,

Answer 1

You want $$ p(D)f=D(D^{2}-D+1)(D-1)f = t+e^{t} $$ The operator $D^{2}$ annihilates $t$ and $(D-1)$ annihilates $e^{t}$. Therefore, $$ D^{3}(D^{2}-D+1)(D-1)^{2}f = 0. $$ Because $D^{2}-D+1=(D-1/2+i\sqrt{3}/2)(D-1/2-i\sqrt{3}/2)$, That gives a solution $$ f = A + Bt + Ct^{2}+Ee^{t/2}\cos(\sqrt{3}t/2)+Fe^{t/2}\sin(\sqrt{3}t/2)+Ge^{t}+Hte^{t}. $$ When you plug back into the original equation, $D(D^{2}-D+1)(D-1)$ annihilates the terms with $A$, $E$, $F$ and $G$. The remaining terms are $$ g = Bt+Ct^{2}+Hte^{t}. $$ Then, \begin{align} p(D)f = p(D)g & = (D^{2}-D+1)(D-1)[D(Bt+Ct^{2})] \\ & +D(D^{2}-D+1)[(D-1)(Hte^{t})] \\ & = (D^{2}-D+1)(D-1)(B+2Ct) \\ & + D(D^{2}-D+1)He^{t} \\ \end{align} Single powers of $D$ annihilate $B$ and higher powers annihilate $2Ct$. Therefore, $$ (D^{2}-D+1)(D-1)(B+2Ct) = (2D-1)(B+2Ct)=(4C-B)-2Ct $$ And, $De^{t}=e^{t}$. Therefore, $$ D(D-D^{2}+1)He^{t} = (1)(1-1+1)He^{t}=He^{t}. $$ Finally, $$ p(D)f = (-B+4C)-2Ct+He^{t} = t+e^{t}\\ \implies -B+4C=0,\;\; C=-1/2,\;\; H =1 \\ \implies B = 4C=-2. $$ The general solution is then $$ f = A-2t-t^{2}/2+Ee^{t/2}\cos(\sqrt{3}t/2)+Fe^{t/2}\sin(\sqrt{3}t/2)+Ge^{t}+te^{t} $$ I'll leave it to you to solve $f'''(2)=f''(2)=f'(2)=1$ and $f(2)=0$ for the remaining constants $A$, $E$, $F$ and $G$.

Answer 2

Yes exactly for finding roots of $p(s)$. In general, one can write the homogeneous part of a linear differential equation with constant coefficients as $$ p(D)x=a_nD^n x+a_{n-1}D^{n-1}x+…+a_1Dx+a_0x=a_nx^{(n)}+a_{n-1}x^{(n-1)}+…+a_1x'+a_0x $$ and if one knows the roots of the polynomial $p(\lambda)=0$ then the linear factorization $$ p(λ)=a_n(λ-λ_1)…(λ-λ_n) $$ translates into $$ p(D)=a_n(D-λ_1)…(D-λ_n) $$ so that you could transform your ODE of order $n$ into a system of order $1$ setting $x=x_1$ and \begin{alignat}{2} (D-λ_n)x_1&=&x_1'-λ_nx_1&=x_2\\ (D-λ_{n-1})x_2&=&x_2'-λ_{n-1}x_2&=x_3\\ &…\\ (D-λ_2)x_{n-1}&=&x_{n-1}'-λ_2x_{n-1}&=x_n\\ (D-λ_1)x_n&=&x_n'-λ_1x_n&=f(t)/a_n \end{alignat} and then solve it backwards.

However, with this form of the inhomogeneity you can use a trial solution that has the same exponential with polynomial factors of the same degree times the monom of degree equal the multiplicity of the exponential factor as root of $p(s)$.

$t+e^t$ has one term with exponential with factor $0$ and one term with exponential factor $1$. Both are single roots of $p(s)$, thus the trial solution is $$ x_p(t)=(A·t^2+B·t)+C·te^t. $$