91Ó°ÊÓ

The given equation is,$y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y=2+6x-5x^2$

Solve for,$y_n$

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y=0$

Solve the above equation,

$$\begin{gathered} D^3+D^2+3D-5=0 \\ {D}^3+2D^2-D^2+5D-2D-5=0 \\ {D}^3+2D^2+5D-D^2-2D-5=0 \\ D\left(D^2+2D+5\right)-1\left(D^2+2D+5\right)=0 \end{gathered}$$

$$\begin{gathered} \left(D-1\right)\left(D^2+2D+5\right)=0 \\ D-1=0or D^2+2D+5=0 \\ D=1,   D=-1+2i,  D=-1-2i \end{gathered}$$

Then, the general solution is$y_n=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)$

$$\begin{gathered} y=y_n+y_p \\ y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2 \end{gathered}$$

Given initial conditions are,$y\left(0\right)=-1,    y\text{'}\left(0\right)=1,    y\text{'}\text{'}\left(0\right)=-3;$

Firstly, solve for,$y\left(0\right)=-1$

One has,$y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2$

Substitute $y\left(0\right)=-1$in the above equation,

$$\begin{gathered} -1=c_1{e}^0+c_2{e}^{-\left(0\right)}\cos \left(0\right)+c_3{e}^{-\left(0\right)}\sin \left(0\right)+{\left(0\right)}^2 \\ -1=c_1+c_2 \\ {c}_1+c_2=-1         ......\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

One has,

$$\begin{gathered} y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2\backslash \mathrm{hfill} \\ y\text{'}=c_1{e}^x+c_2\left(-e^{-x}\cos \left(2x\right)-2e^{-x}\sin \left(2x\right)\right)+c_3\left(-e^{-x}\sin \left(2x\right)+2e^{-x}\cos \left(2x\right)\right)+2x\backslash \mathrm{hfill} \end{gathered}$$

Substitute$y\text{'}\left(0\right)=1$ in the above equation,

$$\begin{gathered} 1=c_1{e}^0+c_2\left(-e^{-\left(0\right)}\cos \left(0\right)-2e^{-\left(0\right)}\sin \left(0\right)\right)+c_3\left(-e^{-\left(0\right)}\sin \left(0\right)+2e^{-\left(0\right)}\cos \left(0\right)\right)+2\left(0\right) \\ 1=c_1+c_2\left(-1\right)+c_3\left(2\right) \\ {c}_1-c_2+2c_3=1         ......\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

One has,

$$\begin{gathered} y\text{'}=c_1{e}^x+c_2\left(-e^{-x}\cos \left(2x\right)-2e^{-x}\sin \left(2x\right)\right)+c_3\left(-e^{-x}\sin \left(2x\right)+2e^{-x}\cos \left(2x\right)\right)+2x\backslash \mathrm{hfill} \\ y\text{'}\text{'}=c_1{e}^x+c_2\left(-3e^{-x}\cos \left(2x\right)+4e^{-x}\sin \left(2x\right)\right)+c_3\left(-3e^{-x}\sin \left(2x\right)-4e^{-x}\cos \left(2x\right)\right)+2\backslash \mathrm{hfill} \end{gathered}$$

Substitute$y\text{'}\text{'}\left(0\right)=-3$ in the above equation,

$$\begin{gathered} -3=c_1{e}^0+c_2\left(-3e^{-\left(0\right)}\cos \left(2\left(0\right)\right)+4e^{-\left(0\right)}\sin \left(2\left(0\right)\right)\right) \\ +c_3\left(-3e^{-\left(0\right)}\sin \left(2\left(0\right)\right)-4e^{-\left(0\right)}\cos \left(2\left(0\right)\right)\right)+2 \\ -3=c_1+c_2\left(-3\right)+c_3\left(-4\right)+2 \\ {c}_1-3c_2-4c_3+2=-3 \\ {c}_1-3c_2-4c_3=-5         ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

Solve the equation (2) and (3),

$$\begin{gathered} 2\left(c_1-c_2+2c_3\right)=1×2 \\ 2c_1-2c_2+4c_3=2 \\ {c}_1-3c_2-4c_3=-5 \\ 3c_1-5c_2=-3=         ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

Now, solve the equation (1) and (4),

$$\begin{gathered}    \left(c_1+c_2=-1\right)5 \\ 5c_1+5c_2=-5 \\  3c_1-5c_2=-3 \\              \stackrel{¯}{8c_1=-8} \\                 c_1=-1 \end{gathered}$$

Substitute the value of $c_1$ in the equation (1),

$$\begin{gathered} c_1+c_2=-1 \\ \left(-1\right)+c_2=-1 \\ {c}_2=0 \end{gathered}$$

Substitute the value of $c_1, c_2$ in the equation (2),

$$\begin{gathered} c_1-c_2+2c_3=1 \\ -1-0+2c_3=1 \\ 2c_3=2 \\ {c}_3=1 \end{gathered}$$

Substitute the value of$c_1, c_2$ and$c_3$ in the general solution.

$$\begin{gathered} y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2 \\ y=\left(-1\right)e^x+\left(0\right)e^{-x}\cos \left(2x\right)+\left(1\right)e^{-x}\sin \left(2x\right)+x^2 \\ y=e^{-x}\sin \left(2x\right)-e^x+x^2 \end{gathered}$$