7.3: Annihilation for Higher Order Equations
- Page ID
- 103559
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Here we consider the constant coefficient nonhomogeneous equation
\[\label{eq:9.3.1} a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=F(x).\]
The theory and methods are the same as in Section 5.5 and will be omitted here, and we will proceed directly to a few examples.
Find the general solution of
\[\label{eq:9.3.2} y'''+y''=e^x\cos x\]
Solution
We first solve \[y'''+y''=0\nonumber\]
Its characteristic equation is
\[r^3+r^2=r^2(r+1)=0\nonumber\]
which gives us our homogeneous solution
\[y_h=c_1+c_2x+c_3e^{-x}.\nonumber\]
Now turning to the nonhomogeneous equation and writing it using differential operator notation we get
\[D^2(D+1)y=e^x\cos x.\nonumber\]
We now annihilate the right hand side using \(((D-1)^2+1)\) to turn it into a new homogeneous equation:
\[((D-1)^2+1)D^2(D+1)y=0\nonumber\]
So, our particular solution takes the form
\[y_p=Ae^x\cos x+Be^x\sin x\nonumber\]
and \(y_p'=Ae^x\cos x-Ae^x\sin x+Be^x\sin x+Be^x\cos x\), \(y_p''=-2Ae^x\sin x+2Be^x\cos x\), and \(y_p'''=-2Ae^x\sin x-2Ae^x\cos x+2Be^x\cos x-2Be^x\sin x\)
Substituting into \ref{eq:9.3.2} gives us
\[(-2Ae^x\sin x-2Ae^x\cos x+2Be^x\cos x-2Be^x\sin x)+(-2Ae^x\sin x+2Be^x\cos x)=e^x\cos x\nonumber\]
This leads to \[(-2A+4B)e^x\cos x+(-4A-2B)e^x\sin x=e^x\cos x.\nonumber\]
So, \(-2A+4B=1\) and \(-4A-2B=0\) giving us
\[A=-{1\over 10}, B={1\over 5}\nonumber\]
and \[y_p=-{1\over 10}e^x\cos x+{1\over 5}e^x\sin x\nonumber\] and the general solution is \[y=c_1+c_2x+c_3e^{-x}-{1\over 10}e^x\cos x+{1\over 5}e^x\sin x.\nonumber\]
Find the general solution of
\[\label{eq:9.3.3} y'''-3y''+3y'-y=e^x-x+16.\]
Solution
We first solve \[y'''-3y''+3y'-y=0\nonumber\]
Its characteristic equation is
\[r^3-3r^2+3r-1=(r-1)^3=0\nonumber\]
which gives us our homogeneous solution
\[y_h=c_1e^x+c_2xe^x+c_3x^2e^x.\nonumber\]
Now turning to the nonhomogeneous equation and writing it using differential operator notation we get
\[(D-1)^3y=e^x-x+16.\nonumber\]
We now annihilate the right hand side using \((D-1)D^2\) to turn it into a new homogeneous equation:
\[(D-1)^4D^2y=0\nonumber\]
Note that while the root \(0\) is occurring for the first and second time, the root \(1\) is occurring for the fourth time. So, our particular solution takes the form
\[y_p=Ax^3e^x+Bx+C\nonumber\]
and \(y_p'=Ax^3e^x+3Ax^2e^x+B\), \(y_p''=Ax^3e^x+6Ax^2e^x+6Axe^x\), and \(y_p'''=Ax^3e^x+9Ax^2e^x+18Axe^x+6Ae^x\)
Substituting into \ref{eq:9.3.3} gives us
\[(Ax^3e^x+9Ax^2e^x+18Axe^x+6Ae^x)-3(Ax^3e^x+6Ax^2e^x+6Axe^x)+3(Ax^3e^x+3Ax^2e^x+B)-(Ax^3e^x+Bx+C)=e^x-x+16\nonumber\]
This leads to \[6Ae^x-Bx+(3B-C)=e^x-x+16\nonumber\]
So, \[A={1\over 6}, B=1, C=-13\nonumber\]
and \[y_p={1\over 6}e^x+x-13\nonumber\] and the general solution is \[y=c_1e^x+c_2xe^x+c_3x^2e^x+{1\over 6}e^x+x-13\nonumber\]
Find the general solution of
\[\label{eq:9.3.4} y'''-4y''+4y'=5x^2-6x+4x^2e^{2x}+3e^{5x}.\]
Solution
We first solve \[y'''-4y''+4y'=0\nonumber\]
Its characteristic equation is
\[r^3-4r^2+4r=r(r-2)^2=0\nonumber\]
which gives us our homogeneous solution
\[y_h=c_1+c_2e^{2x}+c_3xe^{2x}.\nonumber\]
Now turning to the nonhomogeneous equation and writing it using differential operator notation we get
\[D(D-2)^2y=5x^2-6x+4x^2e^{2x}+3e^{5x}.\nonumber\]
We now annihilate the right hand side using \(D^3(D-2)^3(D-5)\) to turn it into a new homogeneous equation:
\[D^4(D-2)^5(D-5)y=0\nonumber\]
Note that while the root \(5\) is occurring for the first time, the root \(0\) is occurring for the second, third, and fourth time and the root \(2\) is occurring for the third, fourth, and fifth time. So, our particular solution takes the form
\[y_p=Ax+Bx^2+Cx^3+De^{5x}+Ex^2e^{2x}+Fx^3e^{2x}+Gx^4e^{2x}\nonumber\]
and \(y_p'=A+2Bx+3Cx^2+5De^{5x}+2Ex^2e^{2x}+2Exe^{2x}+2Fx^3e^{2x}+3Fx^2e^{2x}+2Gx^4e^{2x}+4Gx^3e^{2x}\), \(y_p''=2B+6Cx+25De^{5x}+4Ex^2e^{2x}+8Exe^{2x}+2Ee^{2x}+4Fx^3e^{2x}+12Fx^2e^{2x}+6Fxe^{2x}+4Gx^4e^{2x}+16Gx^3e^{2x}+12Gx^2e^{2x}\), and \(y_p'''=6C+125De^{5x}+8Ex^2e^{2x}+24Ex e^{2x}+12Ee^{2x}+8Fx^3e^{2x}+36Fx^2e^{2x}+36Fxe^{2x}+6Fe^{2x}+8Gx^4e^{2x}+48Gx^3e^{2x}+72Gx^2e^{2x}+24Gxe^{2x}\)
Substituting into \ref{eq:9.3.4} gives us
\[(6C+125De^{5x}+8Ex^2e^{2x}+24Ex e^{2x}+12Ee^{2x}+8Fx^3e^{2x}+36Fx^2e^{2x}+36Fxe^{2x}+6Fe^{2x}+8Gx^4e^{2x}+48Gx^3e^{2x}+72Gx^2e^{2x}+24Gxe^{2x})-4(2B+6Cx+25De^{5x}+4Ex^2e^{2x}+8Exe^{2x}+2Ee^{2x}+4Fx^3e^{2x}+12Fx^2e^{2x}+6Fxe^{2x}+4Gx^4e^{2x}+16Gx^3e^{2x}+12Gx^2e^{2x})+4(A+2Bx+3Cx^2+5De^{5x}+2Ex^2e^{2x}+2Exe^{2x}+2Fx^3e^{2x}+3Fx^2e^{2x}+2Gx^4e^{2x}+4Gx^3e^{2x})=5x^2-6x+4x^2e^{2x}+3e^{5x}.\nonumber\]
This leads to \[(4A-8B+6C)+(8B-24C)x+12Cx^2+45De^{5x}+24Gx^2e^{2x}+(12F+24G)xe^{2x}+(4E+6F)e^{2x}=5x^2-6x+4x^2e^{2x}+3e^{5x}\nonumber\]
So, \[A={3\over 8}, B={1\over 2}, C={5\over 12}, D={1\over 15}, E={1\over 2}, F=-{1\over 3}, G={1\over 6}\nonumber\]
and \[y_p={3\over 8}x+{1\over 2}x^2+{5\over 12}x^3+{1\over 15}e^{5x}+{1\over 2}x^2e^{2x}-{1\over 3}x^3e^{2x}+{1\over 6}x^4e^{2x}\nonumber\] and the general solution is \[y=c_1+c_2e^{2x}+c_3xe^{2x}+{3\over 8}x+{1\over 2}x^2+{5\over 12}x^3+{1\over 15}e^{5x}+{1\over 2}x^2e^{2x}-{1\over 3}x^3e^{2x}+{1\over 6}x^4e^{2x}.\nonumber\]