ODE, An Algebraic Approach (3)

We already know how to solve $P(D)y=0$$P(D)y=0$. Read Math Essays: ODE, An Algebraic Approach (1) (wuyulanliulongblog.blogspot.com)

And for two linear maps $T_1,T_2$$T_1,T_2$, If we have $T_{1}v=u,T_{2}u=0$$T_1 v=u,T_2 u=0$, Then $T_2\circ T_{1}v=0$$T_2\circ T_1 v=0$

And $\{v\in V|T_{1}v=u\}\subseteq Ker{T}_2\circ T_1$$\{v\in V| T_1v=u\}\subseteq Ker T_2\circ T_1$.

That is the basic idea.

For example(to illustrate that and for general, I can not use the best way to solve that),

Example 1.1 $(D^2-4)y=3\cos x$$(D^2-4)y=3\cos x$

We know that $\cos x\in Ker(D^2+1)$$\cos x\in Ker (D^2+1)$

Thus we have $(D^2+1)(D^2-4)y=0$$(D^2+1)(D^2-4)y=0$

Thus we have $(D+i)(D-i)(D+2)(D-2)y=0$$(D+i)(D-i)(D+2)(D-2)y=0$

Therefore the solution of $(D^2-4)y=3\cos x\subseteq Ker(D+i)(D-i)(D+2)(D-2)$$(D^2-4)y=3\cos x\subseteq Ker(D+i)(D-i)(D+2)(D-2)$

Thus $y\in C_1{e}^{ix}+C_2{e}^{-ix}+C_3{e}^{2x}+C_4{e}^{-2x}=(C_1+C_2)\cos x+(C_1-C_2)i\sin x+C_3{e}^{2x}+C_4{e}^{-2x}$$y\in C_1e^{ix}+C_2e^{-ix}+C_3e^{2x}+C_4e^{-2x}=(C_1+C_2)\cos x+(C_1-C_2)i\sin x+ C_3e^{2x}+C_4 e^{-2x}$

And since $(D^2-4)y=3\cos x$$(D^2-4)y=3\cos x$

Thus the coefficient of $\sin x$$\sin x$ should be $0$$0$

Therefore $(D^2-4)2C_1\cos x=3\cos x,-2C_1\cos x-8C_1\cos x=3\cos x$$(D^2-4)2C_1\cos x=3\cos x,-2C_1\cos x-8C_1\cos x=3\cos x$

Thus $C_1=-0.3$$C_1=-0.3$, $y=-0.6\cos x+C_3{e}^{2x}+C_4{e}^{-2x}$$y=-0.6\cos x+C_3 e^{2x}+C_4 e^{-2x}$

Now, Let us consider the limitations of this method.

For $P(D)y=g$$P(D)y=g$, the $g\in KerH(D)$$g\in Ker H(D)$, $H(D)$$H(D)$ is a differential operator polynomial

Remark. ${\lambda }_i,c_i\in \mathbb{C}$$\lambda_i,c_i\in\C$, and $e^{(a+bi)x}=e^a(\cos bx+i\sin bx)$$e^{(a+bi) x}=e^{a}(\cos bx+i\sin bx)$

That means $\begin{array}{c}g=a+e^{{\lambda }_{i}x}(\sum _{i=0}^n{c}_i{x}^i)\end{array}$$g=a+e^{\lambda_ix}(\sum_{i=0}^n c_i x^i)\\$, thus $g$$g$ can be constant, polynomials, $\exp ,\cos x,\sin x,\cosh x,\sinh x$$\exp,\cos x,\sin x,\cosh x,\sinh x$, and product and plus of them...(They can generate a subring(subalgebra actually) of $C^{\mathrm{\infty }}(\mathbb{R})$$C^{\infty}(\mathbb R)$

Example 1.2 $(D^2-9)y=e^x(x^2+3x+1)$$(D^2-9)y=e^x (x^2+3x+1)$

We know that $e^x(x^2+3x+1)\in Ker(D-1{)}^3$$e^x(x^2+3x+1)\in Ker(D-1)^3$

Because $(D-1)(e^{x}f)=({\partial }_1+{\partial }_2-1)(e^{x}f)=(1+{\partial }_2-1)e^{x}f=e^x{f}^{\prime }$$(D-1)(e^xf)=(\part_1+\part_2-1)(e^x f)=(1+\part_2-1)e^x f=e^x f'$

Thus we have $(D-1{)}^3(D^2-9)y=0$$(D-1)^3(D^2-9)y=0$

That is $(D-1{)}^3(D+3)(D-3)y=0$$(D-1)^3(D+3)(D-3)y=0$

Thus we have $y\in e^x(c_1+c_{2}x+c_3{x}^2)+C_4{e}^{-3x}+C_5{e}^{3x}$$y\in e^{ x}(c_1+c_2 x+c_3x^2)+C_4e^{-3x}+C_5e^{3x}$

Recall that $(D^2-9)y=e^x(x^2+3x+1)$$(D^2-9)y=e^x (x^2+3x+1)$

We need to count $(D^2-9)(e^x(c_1+c_{2}x+c_3{x}^2))=e^x(x^2+3x+1)$$(D^2-9)(e^{ x}(c_1+c_2 x+c_3x^2))=e^x (x^2+3x+1)$

$y=\exp (x)\ast (-1/8\ast x^2-7/16\ast x-17/64)$$y=\exp(x)*(-1/8*x^2-7/16*x-17/64)$

Check by Maple.

Summary

As we can see, when $P(D)y=g,g\in KerH(D)$$P(D)y=g,g\in Ker H(D)$, and $gcd(P(D),H(D))=1$$\gcd(P(D),H(D))=1$

The particular solution of $P(D)y=g$$P(D)y=g$ comes from $KerH(D)$$KerH(D)$ since the direct product

$Ker(PH(D)=KerP(D)H(D)=KerP(D)\oplus KerH(D)$$Ker(PH(D)=KerP(D)H(D)=Ker P(D)\oplus KerH(D)$

Thus we can find $y_p$$y_p$ from $KerH(D)$$Ker H(D)$ directly

If $gcd(P(D),H(D))\ne 1$$\gcd(P(D),H(D))\ne 1$

For example, $P(D)=(D-1{)}^2,H(D)=(D-1)(D+1)$$P(D)=(D-1)^2,H(D)=(D-1)(D+1)$, Then we have to find the particular solution in $Ker(D-1{)}^3$$Ker(D-1)^3$

Since we have $Ker(D-1{)}^2(D-1)(D+1)=Ker(D-1{)}^3(D+1)=Ker(D-1{)}^3\oplus Ker(D+1)$$Ker (D-1)^2(D-1)(D+1)=Ker (D-1)^3(D+1)=Ker(D-1)^3\oplus Ker(D+1)$

Example 1.3 $(D^2+3D)y=18$$(D^2+3D)y=18$

The particular solution should come from $Ker{D}^2$$Ker D^2$,

Thus $y_p=c_{1}x$$y_p=c_1 x$, and $y_c=C_2+C_3{e}^{-3x}$$y_c=C_2+C_3e^{-3 x}$

And since $(D^2+3D)y_p=18,3c_1=18,c_1=6$$(D^2+3D)y_p=18,3c_1=18,c_1=6$

And remember, when $a_0=0,$$a_0=0,$ the formal power series has no inverse,

Thus this method shows us a way to solve $\begin{array}{c}\left(\sum _{i=1}^n{D}^i\right)y=P_n(x)\end{array}$$\left(\sum_{i=1}^nD^i\right)y=P_n(x)\\$

Example 1.4

$(D^2+4D)y=1+x+x^2$$(D^2+4D)y=1+x+x^2$

Then $D^3(D^2+4D)=D^4(D+4)y=0$$D^3(D^2+4D)=D^4(D+4)y=0$

Consider the $y_p\in Ker{D}^4$$y_p\in Ker D^4$

$\begin{array}{c}{y}_p=c_3{x}^3+c_2{x}^2+c_{1}x+c_0\end{array}$$y_p= c_3x^3+c_2x^2+c_1x+c_0\\$

$4(3c_3{x}^2+2c_{2}x+c_1)+6c_{3}x+2c_2=x^2+x+1$$4(3c_3x^2+2c_2x+c_1)+6c_3 x+2c_2=x^2+x+1$

Thus $c_3=1/12,8c_2+6c_3=1,4c_1+2c_2=1$$c_3=1/12,8c_2+6c_3=1,4c_1+2c_2=1$

$8c_2=1/2,c_2=1/16,c_1=7/32$$8c_2=1/2,c_2=1/16,c_1=7/32$

Thus $y=1/12x^3+1/16x^2+7/32x+c_0+C{e}^{-4x}$$y=1/12 x^3+ 1/16x^2+7/32x+c_0+Ce^{-4 x}$

Check

$1/2x+1/8+4(1/4x^2+1/8x+7/32)=1/2x+1/8+x^2+1/2x+7/8=x^2+x+1$$1/2 x+1/8+4(1/4 x^2+1/8x+7/32)=1/2x+1/8+x^2+1/2x+7/8=x^2+x+1$