91Ó°ÊÓ

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

Since - 1 and - 5 are homogeneous solutions

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

Hence,

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

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

Hence,

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

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

Hence,

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

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

the corresponding differential equation

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

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

Hence,

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