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Justifying the Method of Undetermined Coefficients

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Feb 24, 2012
The annihilator method can be used to derived the entries in the following table for the method of undetermined coefficients, familiar to all students of ordinary differential equations:

(I)$p_n(x)=a_nx^n+\cdots+a_1x+a_0$$x^sP_n(x)=x^s\left(A_nx^n+\cdots+A_1x+A_0 \right)$
(II)$ae^{\alpha x}$$x^sAe^{\alpha x}$
(III)$a\cos(\beta x)+b\sin(\beta x)$$x^s\left(A\cos(\beta x)+B\sin(\beta x) \right)$
(IV)$p_n(x)e^{\alpha x}$$x^sP_n(x)e^{\alpha x}$
(V)$p_n(x)\cos(\beta x)+q_m\sin(\beta x)$
where $q_m(x)=b_mx^m+\cdots+b_1x+b_0$
$x^s\left(P_N(x)\cos(\beta x)+Q_N(x)\sin(\beta x) \right)$
where $Q_N(x)=B_Nx^N+\cdots+B_1x+B_0$ and $N=\max(n,m)$
(VI)$ae^{\alpha x}\cos(\beta x)+be^{\alpha x}\sin(\beta x)$$x^s\left(Ae^{\alpha x}\cos(\beta x)+Be^{\alpha x}\sin(\beta x) \right)$
(VII)$p_ne^{\alpha x}\cos(\beta x)+q_me^{\alpha x}\sin(\beta x)$$x^se^{\alpha x}\left(P_N(x)\cos(\beta x)+Q_N(x)\sin(\beta x) \right)$
where $N=\max(n,m)$


The non-negative integer $s$ is chosen to be the smallest integer so that no term in the particular solution $y_p(x)$ is a solution to the corresponding homogeneous solution $L[y](x)=0$.

$P_n(x)$ must include all its terms even if $p_n(x)$ has some terms that are zero.

To show this, it suffices to work with type VII functions--that is, functions of the form:

(1) \(\displaystyle g(x)=p_n(x)e^{\alpha x}\cos(\beta x)+q_m(x)e^{\alpha x}\sin(\beta x)\)

where $p_n$ and $q_m$ are polynomials of degrees $n$ and $m$ respectively--since the other types listed in the table are just special cases of (1).

Consider the inhomogeneous equation:

(2) \(\displaystyle L[y](x)=g(x)\)

where $L$ is the linear operator:

(3) \(\displaystyle L\equiv a_nD^{n}+a_{n-1}D^{n-1}+\cdots+a_1D+a_0\)

with $a_n,\,a_{n-1},\,\cdots\,a_0$ constants, and $g(x)$ as given in equation (1). Let $N=\max(n,m)$.

Now, we need to find an annihilator for $g$. If we consider the function:

\(\displaystyle f(x)=e^{\alpha x}\sin(\beta x)\)

we find that:

\(\displaystyle f'(x)=e^{\alpha x}\left(\alpha\sin(\beta x)+\beta\cos(\beta x) \right)\)

\(\displaystyle f''(x)=e^{\alpha x}\left(\left(\alpha^2-\beta^2 \right)\sin(\beta x)+2\alpha\beta\cos(\beta x) \right)\)

If we observe that:

\(\displaystyle f''(x)-2\alpha f'(x)+\left(\alpha^2+\beta^2 \right)f(x)=0\)

then we may state that:

\(\displaystyle D^2-2\alpha D+\alpha^2+\beta^2=(D-\alpha)^2+\beta^2\)

annihilates $f(x)$, and so we conclude that:

\(\displaystyle A\equiv\left((D-\alpha)^2+\beta^2 \right)^{N+1}\)

annihilates $g$.

Now, we need to find the auxiliary equation associated with:

\(\displaystyle AL[y]=0\)

\(\displaystyle \left((D-\alpha)^2+\beta^2 \right)^{N+1}\left(a_nD^n+a_{n-1}D^{n-1}+\cdots+a_0 \right)=0\)

Suppose $0\le s$ is the multiplicity of the roots $\alpha\pm\beta i$ of the auxiliary equation associated with $L[y]=0$, and $r=2_{2s+1}\,\cdots\,r_n$ are the remaining roots, then we have:

(4) \(\displaystyle \left((r-\alpha)^2+\beta^2 \right)^{s+N+1}\left(r-r_{2s+1} \right)\,\cdots\,\left(r-r_n \right)=0\)

Now, as the solution to \(\displaystyle AL[y]=0\) can be written in the form:

\(\displaystyle y(x)=y_h(x)+y_p(x)\)

and we must have:

\(\displaystyle y_h(x)=p_{s+1}(x)e^{\alpha x}\left(\cos(\beta x)+\sin(\beta x) \right)+\sum_{k=2s+1}^n e^{k}\)

then we may conclude that:

\(\displaystyle y_p(x)=x^se^{\alpha x}\left(P_N(x)\cos(\beta x)+Q_N\sin(\beta x) \right)\)

Questions and comments should be posted here:

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