Question

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

$\mathbf{2}{\mathit{y}}^{\mathbf{4}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The general solution of each differential equation is

$y=c_1{e}^{-2x}+c_2{e}^{\frac{x}{2}}+c_3\cos \sqrt{2}x+c_4\sin \sqrt{2}x$

Step-by-step solution

Given Data

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

Given differential equation,

$2y^4+3y\text{'}\text{'}\text{'}+2y\text{'}\text{'}+6y\text{'}-4y=0$

Find auxiliary equation

To find auxiliary equation,

$y=e^{mx}$

$$\begin{gathered} y\text{'}=m{e}^{mx} \\ y\text{'}\text{'}=m^2{e}^{mx} \\ y\text{'}\text{'}\text{'}=m^3{e}^{mx} \\ {y}^4=m^4{e}^{mx} \end{gathered}$$

Substitute the value and find general solution

Substitute$y,y\text{'},y\text{'}\text{'},y\text{'}\text{'}\text{'},y^4$in equation (1)

we get,

$(2m^4+3m^3+2m^2+6m-4)e^{mx}=0$
Since$e^{mx}\ne 0$.

Therefore,

$$\begin{gathered} 2m^4+3m^3+2m^2+6m-4=0 \\ \left(m+2\right)\left(2m-1\right)\left(m^2+2\right)=0 \end{gathered}$$

So,$m=-2,\frac{1}{0},\pm \sqrt{2}i;$

$i\to$imaginary

Thus, the general solution of each differential equation is

$y=c_1{e}^{-2x}+c_2{e}^{\frac{x}{2}}+c_3\cos \sqrt{2}x+c_4\sin \sqrt{2}x$