Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{6}$

Short answer

The general solution is

$y=c_1+c_{2}x+c_3{e}^x-3x^2$

Step-by-step solution

Given Data

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

Given differential equation,

$y\text{'}\text{'}\text{'}-y\text{'}\text{'}=6$

Find auxiliary equation

$y\text{'}\text{'}\text{'}-y\text{'}\text{'}=6$…………(1)
Auxiliary equation of (1) is $f\left(m\right)=m^3-m^2=0$

$$\begin{gathered} \Rightarrow m^2(m-1)=0 \\ \Rightarrow m=0,1 \end{gathered}$$

i.e., roots ore real.

Therefore, of (1) is $y_c=c_1+c_{2}x+c_3{e}^x$.

Let

$$\begin{gathered} y_p=\frac{1}{D^3-D^2}\left(6\right) \\ =\frac{1}{D^2\left(D-1\right)}\left(6\right) \\ =\frac{-1}{D^2}·{\left(1-D\right)}^{-1}\left(6\right) \end{gathered}$$

Find general solution

Simplify,

$$\begin{gathered} =\frac{-1}{D^2}\left[1+D+D^2+\cdots \right] \\ =\frac{-1}{D^2}\left(6+0+\cdots +0\right) \\ =\frac{-1}{D}\int 6dx \\ =\frac{-1}{D}\left(6x\right) \\ =-6\int xdx \\ =-\frac{6x^2}{2} \\ {y}_p=-3x^2 \end{gathered}$$

Thus, the general solution of (1) is $y=y_c+y_p$

$y=c_1+c_{2}x+c_3{e}^x-3x^2$