Question

Use the annihilator method to determine the form of a particular solution for the given equation.

(a)${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$

(b)${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{19}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{20}\mathit{y}\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

(c)${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathbf{3}\mathit{x}$

(d)${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}$

Short answer

1. $y_p=C_0+C_{1}x+C_2{x}^2+C_{3}x{e}^x$
   
2. $y_p=C_{4}x{e}^{-x}$
   
3. $y=C_0+C_{1}x+C_2{x}^2+C_3\cos \sqrt{3}x+C_4\sin \sqrt{3}x+C_{5}x\cos \sqrt{3}x+C_{6}x\sin \sqrt{3}x+C_7\cos 3x+C_8\sin 3x$
   
4. $y_p=C_3\cos x+C_4\sin x+C_{5}x\cos x+C_{6}x\sin x$
   

Step-by-step solution

Determining the form of a particular solution for the given equation

$$\begin{gathered} y^{\text{'}\text{'}}+6y^{\text{'}}+5y=e^{-x}+x^2-1 \\ {e}^{-x}\to D+1 \\ {x}^2-1\to D^3 \\ \left(D^2+6D+5\right)(D+1)D^3=0 \\ (D+1{)}^2(D+5)D^3=0 \\ D=0,0,0,-5,-1,-1 \\ y=C_0+C_{1}x+C_2{x}^2+C_3{e}^{-5x}+C_4{e}^{-x}+C_{5}x{e}^{-x} \end{gathered}$$

Since - 1 and - 5 are homogeneous solutions

$y_p=C_0+C_{1}x+C_2{x}^2+C_{3}x{e}^x$

Hence,

$y_p=C_0+C_{1}x+C_2{x}^2+C_{3}x{e}^x$

Determining the form of a particular solution for the given equation

$$\begin{gathered} y^{\text{'}\text{'}\text{'}}+2y^{\text{'}\text{'}}-19y^{\text{'}}-20y=x{e}^{-x} \\ {e}^{-x}\to D+1 \\ x{e}^{-x}\to (D+1{)}^2 \\ \left(D^3+2D^2-19D-20\right)(D+1{)}^2=0 \\ (D+1)\left(D^2+D-20\right)(D+1{)}^2=0 \\ (D-4)(D+5)(D+1{)}^3=0 \\ D=4,-5,-1,-1 \\ y=C_1{e}^{4x}+C_2{e}^{5x}+C_3{e}^{-x}+C_{4}x{e}^{-x} \\ {y}_p=C_{4}x{e}^{-x} \end{gathered}$$

Hence,

$y_p=C_{4}x{e}^{-x}$

Determining the form of a particular solution for the given equation

$$\begin{gathered} y^4+6y^{\text{'}\text{'}}+9y=x^2-\sin 3x \\ {x}^2\to D^3 \\ \sin 3x\to D^2+9 \\ \left(D^4+6D^2+9\right)D^3\left(D^2+9\right)=0 \\ {D}^3{\left(D^2+3\right)}^2\left(D^2+9\right)=0 \\ \to D=0,0,0,\pm \sqrt{3i},\pm \sqrt{3i},\pm 3i \\ y=C_0+C_{1}x+C_2{x}^2+C_3\cos \sqrt{3}x+C_4\sin \sqrt{3}x \\ +C_{5}x\cos \sqrt{3}x+C_{6}x\sin \sqrt{3}x+C_7\cos 3x+C_8\sin 3x \end{gathered}$$

Hence,

$y=C_0+C_{1}x+C_2{x}^2+C_3\cos \sqrt{3}x+C_4\sin \sqrt{3}x+C_{5}x\cos \sqrt{3}x+C_{6}x\sin \sqrt{3}x+C_7\cos 3x+C_8\sin 3x$

Determining the form of a particular solution for the given equation

$y^{\text{'}\text{'}\text{'}}-y^{\text{'}\text{'}}+2y=x\sin x$

the corresponding differential equation

role="math" localid="1663978603548" $$\begin{gathered} \sin x\to D^2+1 \\ x\sin x\to {\left(D^2+1\right)}^2 \end{gathered}$$

$$\begin{gathered} \left(D^3-D^2+2\right){\left(D^2+1\right)}^2=0 \\ (D+1)\left(D^2-2D+2\right){\left(D^2+1\right)}^2=0 \\ \left[{(D-1)}^2+1\right](D+1){\left(D^2+1\right)}^2=0 \\ \to D=1\pm i,-1,\pm i,\pm i \\ y=\left[C_0\cos x+C_1\sin x\right]e^x+C_2{e}^{-x}+C_3\cos x+C_4\sin x \\ +C_{5}x\cos x+C_{6}x\sin x \\ {y}_p=C_3\cos x+C_4\sin x+C_{5}x\cos x+C_{6}x\sin x \end{gathered}$$

Hence,

$y_p=C_3\cos x+C_4\sin x+C_{5}x\cos x+C_{6}x\sin x$