ODE An Algebraic Approach (4)

We already know two ways to solve certain ordinary differential equations (ODEs) of the form $P(D)y=g$$P(D)y=g$.

One approach is to consider the formal power series inverse,

Math Essays: ODE An Algebraic Approach (2) (wuyulanliulongblog.blogspot.com)

while the other approach is to consider $g\in \text{Ker}H(D)$$g\in \text{Ker} H(D)$.

Math Essays: ODE, An Algebraic Approach (3) (wuyulanliulongblog.blogspot.com)

Each method has its advantages and disadvantages. However, we can actually combine these two methods.

Suppose we have $P(D)y=g$$P(D)y=g$ with $g\in \text{Ker}H(D)$$g\in \text{Ker} H(D)$ and $\text{gcd}(F(D),H(D))=1$$\text{gcd}(F(D),H(D))=1$, where $P(D)=H(D)+F(D)$$P(D)=H(D)+F(D)$.

To find the particular solution, we can simply consider $\begin{array}{c}\frac{1}{F(D)}g\end{array}$$\frac{1}{F(D)}g\\$.

It can be shown that if $y_p$$y_p$ is a particular solution of $P(D)y=g$$P(D)y=g$, then $y_p$$y_p$ is also a particular solution of $F(D)y=g$$F(D)y=g$.

Proof:

$\Rightarrow$$\Rightarrow$

$P(D)y_p=(H(D)+F(D))y_p=F(D)y_p=g$$P(D)y_p=(H(D)+F(D))y_p=F(D)y_p=g$

Since $gcd(P(D),H(D))=1,y_p\in KerH(D)$$\gcd(P(D),H(D))=1,y_p\in Ker H(D)$

$\Leftarrow$$\Leftarrow$

The particular solution of $F(D)y=g$$F(D)y=g$ also comes from $\text{Ker}H(D)$$\text{Ker} H(D)$, since $gcd(H(D,F(D))$$\gcd(H(D,F(D))$

Thus, $F(D)y_p=P(D)y_p=g◻$$F(D)y_p=P(D)y_p=g\quad\square$

This approach can be much simpler than considering $\begin{array}{c}\frac{1}{P(D)}g\end{array}$$\frac{1}{P(D)}g\\$ in some cases.

I observed this when solving the following example: $y^{″}+8y^{\prime }+16y=-16x-1$$y''+8y'+16y=-16x-1$.

We can rewrite it as $(D^2+8D+16)y=-16x-1$$(D^2+8D+16)y=-16x-1$.

Since $\text{gcd}((D+4{)}^2,D^2)=1$$\text{gcd}((D+4)^2,D^2)=1$, we know that the particular solution is in $\text{Ker}{D}^2$$\text{Ker} D^2$.

In this case, you can simply do the following:

$8k+16kx+16b=-16x-1$$8k+16kx+16b=-16x-1$, where $k=-1$$k=-1$ and $-8+16b=-1$$-8+16b=-1$. Solving this yields $\begin{array}{c}b=\frac{7}{16}\end{array}$$b=\frac{7}{16}\\$, so $\begin{array}{c}{y}_p=-x+\frac{7}{16}\end{array}$$y_p=-x+\frac{7}{16}\\$.

Alternatively, you can consider the formal power series inverse:

$\begin{array}{c}\frac{1}{8D+16}(-16x-1)=\frac{1}{16}\frac{1}{1+\frac{D}{2}}(-16x-1)=\frac{1}{16}(1-\frac{D}{2})(-16x-1)=-x+\frac{7}{16}\\ .\end{array}$$\frac{1}{8D+16}(-16x-1)=\frac{1}{16}\frac{1}{1+\frac{D}{2}}(-16x-1)=\frac{1}{16}(1-\frac{D}{2})(-16x-1)=-x+\frac{7}{16}\\.$

Actually we can consider $P(D)=Q(D)H(D)+R(D),H(D)$$P(D)=Q(D)H(D)+R(D),H(D)$ is where the particluar solution comes.

If $gcd(P(D),H(D)=gcd(H(D),R(D))=1$$\gcd(P(D),H(D)=\gcd(H(D),R(D))=1$, then $y_p$$y_p$ is particular solution of $P(D)y=g⟺$$P(D)y=g\iff$ $y_p$$y_p$ is a particluar solution of $R(D)y=g$$R(D)y=g$

But we can observe that $d|P(D)\wedge d|H(D)\Rightarrow d|R(D)$$d|P(D)\wedge d|H(D)\Rightarrow d|R(D)$, the common divsor of $(P(D),H(D)),(H(D),R(D))$$(P(D),H(D)),(H(D),R(D))$ is same.

So we only need $gcd(P(D),H(D))=1$$\gcd(P(D),H(D))=1$.