In Problem 33 to 38, solve the given differential equations by using the principle of superposition [see the solution of equation (6.29)]. For example, in Problem 33, solve three differential equations with right-hand sides equal to the three different brackets. Note that terms with the same exponential factor are kept together; thus, a polynomial of any degree is kept together in one bracket.

${\mathit{y}}^{\mathbf{"}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{9}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{-}\mathbf{6}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$

The differential equation is given as$y^{"}-2y^{\text{'}}=9x{e}^{-x}-6x^2+4e^{2x}$.

Methods used to solve equation are:

a) Successive Integration of Two First-Order Equations

b) Use of Complex Exponentials

c) Exponential Right-Hand Side

c) Method of Undetermined Coefficients

$$\begin{gathered} \left(D-1\right)\left(D+2\right)\mathit{y}\mathbf{=}{\mathit{y}}^{\mathbf{"}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{y} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{p}}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(x^2+1\right) \end{gathered}$$

The equation is

$$\begin{gathered} y^{"}2y^{\text{'}}=9x{e}^{-x}6x^2+4e^{2x} \\ {y}^{"}-2y^{\text{'}}=0 \end{gathered}$$

Let, $D=\frac{d}{dx}$

$$\begin{gathered} y^{"}-2y^{\text{'}}=\frac{d^{2}x}{d{x}^2}-2\frac{dy}{dx} \\ =\frac{d}{dx}\frac{d}{dx}\left(y\right)-2\frac{d}{dx}\left(y\right) \\ =DD\left(y\right)-2D\left(y\right) \\ =D^{2}y-Dy \\ =\left(D^2-2D\right)y \end{gathered}$$

Auxiliary equation implies that

$$\begin{gathered} m^2-2m=0 \\ {m}^2=2m \\ {m}_1=2m \\ {m}_1=\alpha +i\beta =2 \\ {m}_2\alpha =i\beta =0 \end{gathered}$$

Solve further

$$\begin{gathered} y\left(x\right)=\left(c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x}\right) \\ =c_1{e}^{2x}+c_2 \\ {y}_c\left(x\right)=c_1{e}^{2x}+c_2 \\ {y}^{"}-2y^{\text{'}}=-6x^2 \end{gathered}$$

Now, from above

$$\begin{gathered} y_{p1}\left(x\right)=x^3+\frac{3}{2}{x}^2+\frac{3}{2}x \\ {y}^{"}-2y^{\text{'}}=\left(D^2-2D\right)y \\ =9x{e}^{-x} \\ f\left(x\right)=9x{e}^{-x} \end{gathered}$$

Solve further

$$\begin{gathered} y_{p1}^{"}\left(x\right)-2y_{p1}^{\text{'}}\left(x\right)=\left(6ax+2b\right)-2\left(3a{x}^2+2bx+c\right) \\ \Rightarrow -6a{x}^2+\left(6a-4b\right)x+\left(2b-2c\right) \\ =-6x^2 \end{gathered}$$

Equate the coefficients

$$\begin{gathered} a=1 \\ b=\frac{3}{2} \\ c=\frac{3}{2} \end{gathered}$$

Put the values of coefficients in equation

$$\begin{gathered} y_{p1}\left(x\right)=x^3+\frac{3}{2}{x}^2+\frac{3}{2}x \\ {y}^{"}-2y^{\text{'}}=\left(D^2-2D\right)y \\ =9x{e}^{-x} \\ f\left(x\right)=9x{e}^{-x} \end{gathered}$$

Now, solve

$$\begin{gathered} y_{p2}\left(x\right)=\left(dx+g\right)e^{-x} \\ {y}_{p2}^{\text{'}}\left(x\right)=-dx{e}^{-x}+\left(d-g\right)e^{-x} \\ {y}_{p1}^{"}\left(x\right)=dx{e}^{-x}+\left(g-2d\right)e^{-x} \end{gathered}$$

Solve further

$$\begin{gathered} y_{p1}^{"}\left(x\right)-2y_{p1}^{\text{'}}\left(x\right)=\left[dx{e}^{-x}+\left(g-2d\right)e^{-x}\right]-2\left[-dx{e}^{-x}+\left(d-g\right)e^{-x}\right] \\ \Rightarrow 3dx{e}^{-x}+\left[3g-4d\right]e^{-x} \\ =9x{e}^{-x} \end{gathered}$$

Comparing coefficients,

$$\begin{gathered} d=3 \\ g=4 \end{gathered}$$

Put the coefficients in equation

$$\begin{gathered} y_{p2}\left(x\right)=\left(dx+g\right)e^x \\ =\left(3x+4\right)e^x \\ {y}^{"}-2y^{\text{'}}=4r^{2x} \end{gathered}$$

$f\left(x\right)=\frac{1}{2}{e}^x$in the form of $A{e}^{ax}$

$$\begin{gathered} a\ne \alpha =b=2 \\ {y}_{p3}\left(x\right)=Ax{e}^{2x} \\ {y}_{p3}^{\text{'}}\left(x\right)=2Ax{e}^{2x}+A{e}^{2x} \\ {y}_{p3}^{"}\left(x\right)=4Ax{e}^{2x}+4A{e}^{2x} \end{gathered}$$

Put in equation

$$\begin{gathered} \left(D^2-2D\right){\mathit{y}}_{\mathbf{p}\mathbf{3}}\mathbf{=}{\mathit{y}}_{\mathbf{p}\mathbf{3}}^{\mathbf{"}}\mathbf{-}\mathbf{2}{\mathit{y}}_{\mathbf{p}\mathbf{3}}^{\mathbf{\text{'}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(4Ax{e}^{2x}+4Ax{e}^{2x}\right)\mathbf{-}\mathbf{2}\left(2Ax{e}^{2x}+A{e}^{2x}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4}\mathit{A}{\mathit{e}}^{\mathbf{2}\mathbf{x}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{4}\mathit{A}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{=}\mathbf{4}{\mathit{e}}^{\mathbf{x}} \end{gathered}$$

Solve,

$$\begin{gathered} A=2 \\ {y}_{p3}\left(x\right)=2x{e}^{2x} \end{gathered}$$

Solve further,

$$\begin{gathered} y^{"}-2y^{\text{'}}=\left(D^2-2D\right)y \\ =9x{e}^{-x}-6x^2+4e^x \\ y\left(x\right)=y_{gs}+y_{p1}\left(x\right)+y_{p2}\left(x\right)+y_{p3}\left(x\right) \\ =\left[c_1{e}^{2x}+c_2\right]+\left[x^3+\frac{3}{2}{x}^2+\frac{3}{2}x+\left(3x+4\right)e^{-x}+2x{e}^{2x}\right] \end{gathered}$$

Thus, the solution of differential equation is $\left[c_1{e}^{2x}+c_2\right]+\left[x^3+\frac{3}{2}{x}^2+\frac{3}{2}x+\left(3x+4\right)e^{-x}+2x{e}^{2x}\right]$