Question

In Problems 15 to 28 find the general solution of the given higher-order differential equation.

$\mathbf{2}\frac{{\mathbf{d}}^{\mathbf{5}}\mathbf{x}}{\mathbf{d}{\mathbf{s}}^{\mathbf{5}}}\mathbf{-}\mathbf{7}\frac{{\mathbf{d}}^{\mathbf{4}}\mathbf{x}}{\mathbf{d}{\mathbf{s}}^{\mathbf{4}}}\mathbf{+}\mathbf{12}\frac{{\mathbf{d}}^{\mathbf{3}}\mathbf{x}}{\mathbf{d}{\mathbf{s}}^{\mathbf{3}}}\mathbf{+}\mathbf{8}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{s}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1+c_{2}x+c_3{e}^{-x/2}+c_4{e}^{2x}\cos 2x+c_5{e}^{2x}\sin 2x$

Step-by-step solution

Auxiliary equation

From the first equation, find the auxiliary equation.

$$\begin{gathered} 2y^{\left(5\right)}-7y^{\left(4\right)}+12y^{\text{'}\text{'}}+8y^{\text{'}\text{'}}=0 \\ 2m^5-7m^4+12m^3+8m^2 \end{gathered}$$

Factor out an m².

$m^2\left(2m^3-7m^2+12m+8\right)$

Determine the general equation of the higher – order differential equation

Use synthetic division to factor the obtained auxiliary equation..

To find the possible zeroes (the "divisors" we'll use in the synthetic division), take the factors of the last coefficient in the auxiliary equation and divide them by the factors of the first coefficient in the auxiliary equation.

Factors of $8:\pm 1,\pm 2,\pm 4,\pm 8$

Factors of $2:\pm 1,\pm 2$

Possible Zeroes: $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1},\pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{4}{2},\pm \frac{8}{2}$

Therefore, the possible zeroes are:

Make your best try at which zero you think will work. -1/2 is the one we'll choose.

$2x^3-7x^2+12x+8$

The previous step tells us that -1/2 is a zero, which means $\left(m+\frac{1}{2}\right)$ is a factor, and we know that m² is also part of the equation from step 1.

The numbers under the synthetic division bar tells us the other factor.

$$\begin{gathered} m^2\left(m+\frac{1}{2}\right)\left(2m^2-8m+16\right)=0 \\ {m}_1=0 \end{gathered}$$

With a multiplicity of two

$m_2=-\frac{1}{2}$

Factor $\left(2m^2-8m+16\right)$using the quadratic equation,

$$\begin{gathered} m_3=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ {m}_3=\frac{8\pm \sqrt{64-4\left(2\right)\left(16\right)}}{4} \\ {m}_3=2\pm 2i \end{gathered}$$

Therefore,

$\alpha =2,\beta =2$

Values for $m_1,m_2,\alpha and\beta$

With a multiplicity of two

$$\begin{gathered} m_2=-\frac{1}{2} \\ \alpha =2,\beta =2 \end{gathered}$$

Plug in values for

$y=c_1+c_{2}x+c_3{e}^{-x/2}+c_4{e}^{2x}\cos 2x+c_5{e}^{2x}\sin 2x$

Hence the general solution is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}\mathbf{+}{\mathit{c}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{5}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{x}$