Question

Solve the following differential equations by the methods discussed above and compare computer solutions.

$y^{\text{'}\text{'}}-2y^{\text{'}}=0$

Short answer

The solution is$y=c_1+c_2{e}^{2x}$

Step-by-step solution

Given information

A differential equation is given as $y"-2y\text{'}=0$.

Differential equation.

A differential equation of the form $(D-a)(D-b)y=0,a\ne b$has general solution

$y=c_1{e}^{ax}+c_2{e}^{bx}.$

Find the solution of the given differential equation.

Suppose that $D=\frac{d}{dx}$.

Then

$$\begin{gathered} Dy=\frac{dy}{dx} \\ Dy=y^{\text{'}} \\ {D}^{2}y=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\ \frac{d^{2}y}{d{x}^2}=y^{\text{'}\text{'}} \end{gathered}$$

Given differential equation is $y^{\text{'}\text{'}}-2y^{\text{'}}=0$.

$$\begin{gathered} D^{2}y-2Dy=0 \\ \left(D^2-2D\right)y=0 \end{gathered}$$

Auxiliary equation is-

$$\begin{gathered} D^2-2D=0 \\ D(D-2)=0 \\ D=0 \\ D=2 \end{gathered}$$

These are unequal roots$\Rightarrow D=0,D=2$

Differential equation becomes

$\Rightarrow (D-0)[D-2]y=0$

These are separable equation with solution

$y=c_1{e}^{0x}\text{and}y=c_2{e}^{2x}$

The general solution will be given by

$$\begin{gathered} y=c_1{e}^{0x}+c_2{e}^{2x} \\ \Rightarrow y=c_1+c_2{e}^{2x} \end{gathered}$$