Question

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

$\left(2D^2+D-1\right)y=0$

Short answer

The solution is$y=c_1{e}^{\frac{1}{2}x}+c_2{e}^{-x}$

Step-by-step solution

Given information

A differential equation is given as $4y"+12y\text{'}+9=0$.

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 $\left(2D^2+D-1\right)y=0$.

Auxiliary equation is-

$$\begin{gathered} 2D^2+D-1=0 \\ 2D^2+2D-D-1=0 \\ (2D-1)(D+1)=0 \end{gathered}$$

These are unequal roots$\Rightarrow D=\frac{1}{2},D=-1$

Differential equation becomes

$\Rightarrow \left(D-\frac{1}{2}\right)[D-(-1\left)\right]y=0$

These are separable equation with solution

$y=c_1{e}^{\frac{1}{2}x}\text{and}y=c_2{e}^{-1x}$

The general solution will be given by $y=c_1{e}^{\frac{1}{2}x}+c_2{e}^{-x}$.