Question

In Problems 17-32 use the procedures developed in this chapter to find the general solution of each differential equation.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{8}\mathbf{+}\mathbf{2}\mathbf{sinx}$

Short answer

The general solution of each differential equation is$y=c_1+c_2{e}^{6x}+c_3{e}^{-x}-\frac{4}{3}x+\frac{1}{74}\left(5\sin x+7\cos x\right)$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

$y\text{'}\text{'}\text{'}-5y\text{'}\text{'}+6y\text{'}=8+2\sin x$

In operator form,

$(D^3-5D^2-6D)=8+2\sin x$

Write the characteristics equation

Characteristic equation,

$$\begin{gathered} D^3-5D^2-6D=0D \\ D·\left(D^2-SD-6\right)=0 \\ D(D^2+D-6D-6)=0 \\ D\left(D\right(D+1)-6(D+1\left)\right)=0 \\ D(D-6)(D+1)=0 \end{gathered}$$

Where,

$D=0,D=6,D=-1$

So,

$y_c=c_1+c_2{e}^{6x}+c_3{e}^{-x}$

Simplify the final step

$$\begin{gathered} y_c=\frac{8e^{0x}}{D\left(D^2-5D-6\right)}+\frac{2\sin x}{\left(D^3-5D^2-6D\right)} \\ =\frac{8e^{0x}}{-6·D}+\frac{2\sin x}{D\left(-1^2\right)-5\left(-1^2\right)-6D} \\ =\frac{-4}{3}\int dx+\frac{2\sin x}{-D+5-6D} \\ =-\frac{4}{3}x+\frac{2\sin x}{5-7D} \\ =\frac{-4}{3}+\frac{\left(5+7D\right)}{\left(25-49D^2\right)}\sin x \\ =\frac{-4}{3}x+\frac{\left(5+7D\right)\sin x}{25+49} \end{gathered}$$

$y_p=\frac{-4}{3}x+\frac{1}{74}\left(5\sin x+7\cos x\right)$

$y=c_1+c_2{e}^{6x}+c_3{e}^{-x}-\frac{4}{3}x+\frac{1}{74}(5\sin x+7\cos x)$

Thus, the general solution of each differential equation is

$y=c_1+c_2{e}^{\left(6x\right)}+c_3{e}^{-x}-\frac{4}{3}x+\frac{1}{74}(5\sin x+7\cos x)$