Solve

Solveby the method used in solving,for the following three cases, to obtain the result.

(a) cis not equal to eitheror b;

(b);$\mathit{a}\mathbf{\ne }\mathit{b}\mathbf{,}\mathbf{}\mathit{c}\mathbf{=}\mathit{a}\mathbf{;}$

(c).$\mathbf{a}\mathbf{=}\mathbf{b}\mathbf{=}\mathbf{c}$

The differential equation is,

$$\begin{gathered} \left(D-a\right)\left(D-b\right)\mathbf{y}\mathbf{=}\mathbf{F}\left(x\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{ke}}^{\mathbf{cx}} \end{gathered}$$

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.

$\left(D-1\right)\left(D+2\right)\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}$

(a)

For the case when $\mathrm{c}\ne \mathrm{a}\ne \mathrm{b}$ start by assuming $\mathrm{y}=\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}$, therefore,

$$\begin{gathered} \left(\mathrm{D}-\mathrm{a}\right)\mathrm{u}={\mathrm{ke}}^{\mathrm{cx}} \\ \mathrm{u}-\mathrm{au}={\mathrm{ke}}^{\mathrm{cx}} \end{gathered}$$

This has become first order differential equation which can be solved using eq.(3.4) and eq.(3.9), that is

$$\begin{gathered} \mathrm{l}=-\int \mathrm{adx} \\ =-\mathrm{ax} \\ \Rightarrow {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{ax}} \\ {\mathrm{ue}}^{\mathrm{l}}=\int \left({\mathrm{ke}}^{\mathrm{cx}}\right){\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx}\Rightarrow \frac{\mathrm{k}}{\mathrm{c}-\mathrm{a}}{\mathrm{e}}^{\left(\mathrm{c}-\mathrm{a}\right)\mathrm{x}} \\ \mathrm{u}=\frac{\mathrm{k}}{\mathrm{c}-\mathrm{a}}{\mathrm{e}}^{\mathrm{cx}} \end{gathered}$$

Substitute back$\mathrm{u}=\frac{\mathrm{k}}{\mathrm{c}-\mathrm{a}}{\mathrm{e}}^{\mathrm{cx}}\mathrm{into}\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}=\mathrm{u}$ into , therefore,

$\mathrm{y}-\mathrm{by}=\frac{\mathrm{k}}{\mathrm{c}-\mathrm{a}}{\mathrm{e}}^{\mathrm{cx}}$

Again, this has become first order differential equation which can be solved using eq. (3.4) and eq.(3.9), that is

$$\begin{gathered} \mathrm{l}=\int -\int \mathrm{bdx} \\ =-\mathrm{bx} \\ \Rightarrow {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{bx}} \\ {\mathrm{ye}}^{\mathrm{l}}=\int \left(\frac{\mathrm{k}}{\mathrm{c}-\mathrm{a}}{\mathrm{e}}^{\mathrm{cx}}\right){\mathrm{e}}^{-\mathrm{bx}}\mathrm{dx} \\ =\frac{\mathrm{k}}{\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{b}\right)}{\mathrm{e}}^{\left(\mathrm{c}-\mathrm{b}\right)\mathrm{x}} \\ \mathrm{y}=\frac{\mathrm{k}}{\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{b}\right)}{\mathrm{e}}^{\mathrm{cx}} \end{gathered}$$

(b

For the case when$\mathrm{c}=\mathrm{a}\ne \mathrm{b}$, let $\mathrm{u}=\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}$, therefore,

$$\begin{gathered} \left(\mathrm{D}-\mathrm{a}\right)\mathrm{u}={\mathrm{ke}}^{\mathrm{cx}} \\ \mathrm{u}-\mathrm{au}={\mathrm{ke}}^{\mathrm{cx}} \\ ={\mathrm{ke}}^{\mathrm{ax}} \end{gathered}$$

This is first order linear differential equation that can be solved using eq.(3.4) and

$$\begin{gathered} \mathrm{l}=-\int \mathrm{adx} \\ =-\mathrm{ax} \\ {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{ax}} \end{gathered}$$

$$\begin{gathered} {\mathrm{ue}}^{\mathrm{l}}=\int \left({\mathrm{ke}}^{\mathrm{ax}}\right){\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx} \\ =\mathrm{kx} \\ \mathrm{u}={\mathrm{kxe}}^{\mathrm{ax}} \end{gathered}$$

Substitute $\mathrm{u}={\mathrm{kxe}}^{\mathrm{ax}}$into$\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}=\mathrm{u}$ and obtain the equation as,

$\mathrm{y}-\mathrm{by}={\mathrm{kxe}}^{\mathrm{ax}}$

Which is again a first order linear differential equations thatcan be solved using eq.(3.4) and eq.(3.9), that is

$$\begin{gathered} \mathrm{l}=-\int \mathrm{bdx} \\ =-\mathrm{bx} \\ {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{bx}} \\ {\mathrm{ye}}^{\mathrm{l}}=\int \left({\mathrm{kxe}}^{-\mathrm{ax}}\right){\mathrm{e}}^{-\mathrm{bx}}\mathrm{dx} \\ =\mathrm{k}\left[\frac{\mathrm{x}}{\mathrm{a}-\mathrm{b}}-\frac{1}{{\left(\mathrm{a}-\mathrm{b}\right)}^2}\right]{\mathrm{e}}^{\left(\mathrm{a}-\mathrm{b}\right)\mathrm{x}} \\ \mathrm{y}=\mathrm{k}\left[\frac{\mathrm{x}}{\mathrm{a}-\mathrm{b}}-\frac{1}{{\left(\mathrm{a}-\mathrm{b}\right)}^2}\right]{\mathrm{e}}^{\mathrm{ax}} \end{gathered}$$

(c)

For the case $\mathrm{c}=\mathrm{a}=\mathrm{b}$,let $\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}=\mathrm{u}$, therefore,

$$\begin{gathered} \left(\mathrm{D}-\mathrm{a}\right)\mathrm{u}={\mathrm{ke}}^{\mathrm{cx}} \\ ={\mathrm{ke}}^{\mathrm{ax}} \\ \mathrm{u}-\mathrm{au}={\mathrm{ke}}^{\mathrm{ax}} \end{gathered}$$

The differential equation has become first order linear differential equation that can be solved using eq.(3.4) and eq.(3.9), that is

$$\begin{gathered} \mathrm{l}=\int \left(-\mathrm{a}\right)\mathrm{dx}=-\mathrm{ax} \\ {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{ax}} \\ {\mathrm{ue}}^{\mathrm{l}}=\int \left({\mathrm{ke}}^{\mathrm{ax}}\right){\mathrm{e}}^{-\mathrm{ax}}\mathrm{dx} \\ =\mathrm{kx} \\ \mathrm{u}={\mathrm{kxe}}^{\mathrm{ax}} \end{gathered}$$

Substitute $\mathrm{u}={\mathrm{kxe}}^{\mathrm{ax}}$into $\left(\mathrm{D}-\mathrm{b}\right)\mathrm{y}=\mathrm{u}$and obtain the equation as,

$\mathrm{y}-\mathrm{ay}={\mathrm{kxe}}^{\mathrm{ax}}$

Which is again a first order linear differential equation that can be solved using the same equations

$$\begin{gathered} \mathrm{l}=-\int \mathrm{adx} \\ =-\mathrm{ax} \\ {\mathrm{e}}^{\mathrm{l}}={\mathrm{e}}^{-\mathrm{ax}} \end{gathered}$$

Substitute the value

$$\begin{gathered} y{e}^l=\int \left(kx{e}^{ax}\right)e^{-ax}dx \\ =\frac{k}{2}{x}^2 \\ y=\frac{k}{2}{x}^2{e}^{ax} \end{gathered}$$