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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:

Notes:

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:

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:

where $L$ is the linear operator:

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:

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:

http://mathhelpboards.com/commentar...ng-method-undetermined-coefficients-4840.html

Type | $g(x)$ | $y_p(x)$ |

(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 $N=\max(n,m)$ |

Notes:

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:

http://mathhelpboards.com/commentar...ng-method-undetermined-coefficients-4840.html

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