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.

$\mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{}\mathit{x}$

The differential equation is given as$\left(D^2-1\right)y=\sin hx$.

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

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

The equation is

$$\begin{gathered} \left(D^2-1\right)y=\sin h\left(x\right) \\ \left(D^2-1\right)=0 \end{gathered}$$

Auxiliary equation implies that

$$\begin{gathered} m^2-1=0 \\ {m}^2=1 \\ m=\pm 1 \\ y\left(x\right)=c_1{e}^{m_{1}x}+c_2{e}^{m_{2}x} \\ =c_1{e}^x+c_2{e}^{-x} \end{gathered}$$

Solve further

$$\begin{gathered} y_0\left(x\right)=c_1{e}^x+c_2{e}^x \\ \sin h\left(ax\right)=\frac{e^{ax}-e^{ax}}{2} \end{gathered}$$

Solve equation,

$$\begin{gathered} \left(D^2-1\right)=\sin h\left(x\right) \\ =\frac{e^x-e^{-x}}{2} \\ =\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x} \\ \left(D-1\right)\left(D+1\right)y=\frac{1}{2}y=\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x} \\ \left(D-1\right)\left(D+1\right)y=\frac{1}{2}y=\frac{1}{2}{e}^x \end{gathered}$$

The conditions are

$$\begin{gathered} \propto =1 \\ a=1 \\ b=-1 \\ a=\propto \ne b \\ {y}_{pl}\left(x\right)=Ax{e}^x \end{gathered}$$

Solving further

$$\begin{gathered} y_{p1}\left(x\right)=Ax{e}^x+A{e}^x \\ {y}_{p1}^{"}\left(x\right)=Ax{e}^x+2A{e}^x \\ \left(D-1\right)\left(D+1\right)y=y_{p1}^{"}\left(x\right)-y_{p1}\left(x\right) \\ =\left(Ax{e}^x+2A{e}^x\right) \\ =2A{e}^x \end{gathered}$$

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

$$\begin{gathered} 2A{e}^x=\frac{1}{2}{e}^x \\ A=\frac{1}{4} \\ {y}_{p1}\left(x\right)=\frac{1}{4}x{e}^x \\ \left(D-1\right)\left(D+1\right)y=-\frac{1}{2}{e}^{-x} \end{gathered}$$

The conditions are

$$\begin{gathered} \alpha =-1 \\ a=1 \\ b=-1 \\ b=\alpha \ne a \end{gathered}$$

Solving further

$$\begin{gathered} y_{p2}\left(x\right)=Bx{e}^{-x} \\ {y}_{p2}\left(x\right)=-Bxa-B{e}^{-x} \\ {y}_{p2}^{"}\left(x\right)=Bx{e}^{-x}-2B{e}^{-x} \\ \left(D-1\right)\left(D+1\right)y=y_{p2}^{"}\left(x\right)-y_{p2}\left(x\right) \\ =\left(Bx{e}^{-x}-2B{e}^{-x}\right)-Bx{e}^{-x} \\ =-2B{e}^{-x} \end{gathered}$$

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

$$\begin{gathered} -2B{e}^{-x}=-\frac{1}{2}{e}^{-x} \\ B=\frac{1}{4} \\ {y}_{p2}\left(x\right)=\frac{1}{4}x{e}^{-x} \\ \left(D^2-1\right)y=\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x} \end{gathered}$$

Solving further

$$\begin{gathered} y_p\left(x\right)=y_{p1}\left(x\right)+y_{p2}\left(x\right) \\ =\frac{1}{4}x{e}^x+\frac{1}{4}x{e}^{-x} \\ =\frac{1}{4}x\left[e^x-e^{-x}\right] \\ y\left(x\right)=c_1{e}^x+c_2{e}^{-x}+\frac{1}{4}x\left[e^x-e^{-x}\right] \end{gathered}$$

Thus, the solution of given differential equation is $y\left(x\right)=c_1{e}^x+x_2{e}^{-x}+\frac{1}{4}x\left[e^x-e^{-x}\right]$.