(a) Show that

$\left(D-a\right){\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\left(c-a\right){\mathbf{e}}^{\mathbf{cx}}\mathbf{;}\left(D^2+5D-3\right){\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\left(c^2+5c-3\right){\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\mathbf{L}\left(D\right){\mathbf{e}}^{\mathbf{cx}}\mathbf{,}$where $\mathbf{L}\left(D\right)$is any polynomial in $\mathbf{D}\mathbf{;}\left(D-c\right){\mathbf{xe}}^{\mathbf{cx}}\mathbf{=}{\mathbf{e}}^{\mathbf{cx}}\mathbf{;}\mathbf{}{\left(D-c\right)}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{cx}}$.

(b) Define the expression $\mathbf{y}\mathbf{=}\left[1/L\left(D\right)\right]\mathbf{u}\left(x\right)$to mean a solution of the differential equation $\mathbf{L}\left(D\right)\mathbf{y}\mathbf{=}\mathbf{u}$.

Using part (a), show that;

localid="1659340707727" $$\begin{gathered} \frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{a}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{cx}}}{\mathbf{c}\mathbf{-}\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{c}\mathbf{\ne }\mathbf{a}\mathbf{;}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{D}\mathbf{-}\mathbf{3}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{cx}}}{{\mathbf{c}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{c}\mathbf{-}\mathbf{3}}\mathbf{;}\frac{\mathbf{1}}{\mathbf{L}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{cx}}}{\mathbf{L}\mathbf{\left(}\mathbf{c}\mathbf{\right)}}\mathbf{,}\mathbf{}\mathbf{L}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{\ne }\mathbf{0}\mathbf{;} \\ \frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{a}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}{\mathbf{xe}}^{\mathbf{cx}}\mathbf{;}\frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{a}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{.} \end{gathered}$$

(c) The expressionsin (b) are called inverse operators. They can be used to find particular solutions of differential equations. As an example consider localid="1659340713408" $$\begin{gathered} \mathbf{(}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{2}\mathbf{x}\mathbf{,}} \\ \mathbf{y}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}}{\mathbf{e}}^{\mathbf{2}\mathbf{x}\mathbf{,}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{x}}}{{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{-}\mathbf{2}}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{2}\mathbf{x}}}{\mathbf{4}} \end{gathered}$$,

Using inverse operators, find particular solutions of Problems 4 to 20. Be careful to use parts 4 or 5 of (b) ifis a root of the auxiliary equation. For example,$\frac{\mathbf{1}}{\left(D-a\right)\left(D-c\right)}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{a}}\frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{a}}{\mathbf{e}}^{\mathbf{cx}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{D}\mathbf{-}\mathbf{c}}\frac{{\mathbf{e}}^{\mathbf{cx}}}{\mathbf{c}\mathbf{-}\mathbf{a}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{ex}}}{\mathbf{c}\mathbf{-}\mathbf{a}}\mathbf{.}\mathbf{,}$

It is given in the question that $\mathrm{y}=\left[1/\mathrm{L}\left(\mathrm{D}\right)\right]\mathrm{u}\left(\mathrm{x}\right)$is a solution of the differential equation $\mathrm{L}\left(\mathrm{D}\right)\mathrm{y}=\mathrm{u}$.

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

a)

Recall that D stands for $\frac{\mathrm{d}}{\mathrm{dx}}$.

Then,

$$\begin{gathered} \left(D-a\right)e^{cx}=D{e}^{cx}-a{e}^{cx} \\ =c{e}^{cx}-a{e}^{cx}\left(D-a\right)e^{cx} \\ =\left(c-a\right)e^{cx}\left(D^2+5D-3\right)e^{cx} \\ =D^2{e}^{cx}+5D{e}^{cx}-3e^{cx} \\ =c^2{e}^{cx}+5c{e}^{cx}-3e^{cx} \\ =\left(c^2+5c-3\right)e^{cx} \end{gathered}$$

Similarly, $\mathrm{L}\left(\mathrm{D}\right){\mathrm{e}}^{\mathrm{cx}}=\mathrm{L}\left(\mathrm{c}\right){\mathrm{e}}^{\mathrm{cx}}$, where $\mathrm{L}\left(\mathrm{D}\right)$is any polynomial D.

Now,

localid="1659340724292" $$\begin{gathered} \left(D-c\right)x{e}^{cx}=Dx{e}^{cx}-cx{e}^{cx} \\ =\left(e^{cx}-xc{e}^{cx}\right)-cx{e}^{{}^{cx}} \\ =e^{{}^{cx}} \\ {\left(D-c\right)}^2{x}^2{e}^{cx}=\left(D^2-2Dc+c^2\right)x^2{e}^{cx} \\ =D^2\left(x^2{e}^{cx}\right)-2D\left(c{x}^2{e}^{cx}\right)+c^2{x}^2{e}^{cx} \\ =D\left(2x{e}^{cx}+x^{2}c{e}^{cx}\right)-2c\left(2x{e}^{cx}+x^{2}c{e}^{cx}\right)+c^2{x}^2{e}^{cx} \\ =\left(2x{e}^{cx}+2xc{e}^{cx}+2xc{e}^{cx}+x^2{c}^2{e}^{cx}\right)-4cx{e}^{cx}-2c^2{x}^2{e}^{cx}+c^2{x}^2{e}^{cx} \\ =2x{e}^{cx}+4xc{e}^{cx}+x^2{c}^2{e}^{cx}-4cx{e}^{cx}-2c^2{x}^2{e}^{cx}+c^2{x}^2{e}^{cx} \\ =2e^{cx} \end{gathered}$$

b)

The solution of this equation can be found as follows:

$$\begin{gathered} b) \\ Thesolutionofthisequationcanbefoundasfollows: \\ \left(D-a\right)e^{cx}=\left(c-a\right)e^{cx} \\ \frac{1}{\left(D-a\right)}{e}^{cx}=\frac{1}{c-a}{e}^{cx}\left[ifc\ne a\right] \\ \left(D^2+5D-3\right)e^{cx}=\left(c^2+5c-3\right)e^{cx} \\ \frac{1}{\left(D^2+5D-3\right)}{e}^{cx}=\frac{1}{\left(c^2+5c-3\right)}{e}^{cx} \end{gathered}$$

Solve the problem further

$$\begin{gathered} L\left(D\right)e^{cx}=L\left(c\right)e^{cx} \\ \frac{1}{L\left(D\right)}{e}^{cx}=\frac{1}{L\left(c\right)}{e}^{cx}\left[ifL\left(c\right)\ne 0\right]\left(D-c\right)x{e}^{cx}=e^{cx} \\ \frac{1}{\left(D-c\right)}{e}^{cx}=x{e}^{cx} \end{gathered}$$

Proceed further

$$\begin{gathered} {\left(\mathrm{D}-\mathrm{c}\right)}^2{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cx}}=2{\mathrm{e}}^{\mathrm{cx}} \\ \frac{1}{{\left(\mathrm{D}-\mathrm{c}\right)}^2}{\mathrm{e}}^{\mathrm{cx}}=\frac{1}{2}{\mathrm{x}}^2{\mathrm{e}}^{\mathrm{cx}} \end{gathered}$$

c)

Consider the equation$\left(\mathrm{D}+1\right)\left(\mathrm{D}-3\right)\mathrm{y}=24{\mathrm{e}}^{-3\mathrm{x}}$

The solution of this equation can be found as follows:

$$\begin{gathered} \left(\mathrm{D}+1\right)\left(\mathrm{D}-3\right)\mathrm{y}=24{\mathrm{e}}^{-3\mathrm{x}} \\ \mathrm{y}=\frac{1}{\left(\mathrm{D}+1\right)\left(\mathrm{D}-3\right)}24{\mathrm{e}}^{-3\mathrm{x}} \\ =\frac{1}{\left({\mathrm{D}}^2-2\mathrm{D}-3\right)}{\mathrm{e}}^{-3\mathrm{x}} \\ =\frac{1}{{\left(-3\right)}^2-2\left(-3\right)-3}{\mathrm{e}}^{-3\mathrm{x}}\left[\mathrm{Since}\frac{1}{\mathrm{L}\left(\mathrm{D}\right)}{\mathrm{e}}^{\mathrm{cx}}=\frac{1}{\mathrm{L}\left(\mathrm{D}\right)}{\mathrm{e}}^{\mathrm{cx}},\mathrm{if}\mathrm{L}\left(\mathrm{c}\right)\ne 0\right] \end{gathered}$$

Solve the equation further

$$\begin{gathered} =\frac{1}{9+6-3}{e}^{-3x} \\ =\frac{e^{-3x}}{12} \end{gathered}$$

Similarly, the solution of the other problems can be found using the rules found in part (a) and part (b).