Question

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

$2y"+2y^{\text{'}}+3y=0$

Short answer

The general solution ofdifferential equation is$y=k_1{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{5}}{2}x+k_2{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{5}}{2}x$

Step-by-step solution

Define

The second definition — and the one which you'll see much more often—states that a differential equation (of any order) is homogeneous if once all the terms involving the unknown function are collected together on one side of the equation, the other side is identically zero.

Solve the equation.

We have the second order homogeneous differential equation

$2y"+2y\text{'}+3y=0$

and we have to solve it by assuming that$y=e^{mx}$is a solution of the given D.E, then follow this technique:

Differentiate this assumption with respect to$x$, then we have

$$\begin{gathered} y\text{'}=m{e}^{mx} \\ y"=m^2{e}^{mx} \end{gathered}$$

After that, substitute with $y\text{'}$and $y"$shown in equations and (3) and our assumption $y=e^{mx}$into equation (1), then we obtain

$$\begin{gathered} 2m^2{e}^{mx}+2m{e}^{mx}+3e^{mx}=0 \\ (2m^2+2m+3)e^{mx}=0 \end{gathered}$$

Simplify the equation.

Since$e^{mx}$can not be 0 , then we have the auxiliary equation as

$2m^2+2m+3=0$

Then we have its roots as

$m_{1,2}=\frac{-2\pm \sqrt{4-4\times 2\times 3}}{2\times 2}$

$=\frac{-2\pm \sqrt{-20}}{4}$

$=-\frac{1}{2}\pm \frac{\sqrt{5}}{2}i$

which are conjugate complex
Then we can obtain the solution of the given D.E $2y"+2y\text{'}+3y=0$as

$$\begin{gathered} y=c_1{e}^{\left(-\frac{1}{2}+\frac{\sqrt{5}}{2}i\right)x}+c_2{e}^{\left(-\frac{1}{2}-\frac{\sqrt{5}}{2}i\right)x} \\ =c_1{e}^{-\frac{1}{2}x}{e}^{\left(\frac{\sqrt{5}}{2}i\right)x}+c_2{e}^{-\frac{1}{2}x}{e}^{\left(\frac{\sqrt{5}}{2}i\right)x} \\ =e^{-\frac{1}{2}x}\left[c_1{e}^{\left(\frac{\sqrt{5}}{2}i\right)x}+c_2{e}^{\left(\frac{\sqrt{5}}{2}i\right)x}\right] \\ =e^{-\frac{1}{2}x}\left[c_1\left(\cos \frac{\sqrt{5}}{2}x+\sin \frac{\sqrt{5}}{2}x\right)+c_2\left(\cos \frac{\sqrt{5}}{2}x-\sin \frac{\sqrt{5}}{2}x\right)\right] \\ =e^{-\frac{1}{2}x}\left[(c_1+c_2)\cos \frac{\sqrt{5}}{2}x+(c_1-c_2)\sin \frac{\sqrt{5}}{2}x\right] \\ =k_1{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{5}}{2}x+k_2{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{5}}{2}x \end{gathered}$$

where we have the arbitrary constants$k_1=c_1+c_2$and$k_2=c_1-c_2$

.

The general solution of differential equation is

$y=k_1{e}^{-\frac{1}{2}x}\cos \frac{\sqrt{5}}{2}x+k_2{e}^{-\frac{1}{2}x}\sin \frac{\sqrt{5}}{2}x$