Question

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

${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1+c_{2}x+c_3{e}^{-x/2}\cos \frac{\sqrt{3}}{2}x+c_4{e}^{-x/2}\sin \frac{\sqrt{3}}{2}x$

Step-by-step solution

Determine the general equation of the higher – order differential equation

Since the given equation is

$y^{\left(4\right)}+y\text{'}\text{'}\text{'}+y\text{'}\text{'}=0$

The proper auxiliary equation is

$m^4+m^3+m^2=0$

Solve the equation: $m^4+m^3+m^2=0$(Factor out m²)

$m^2\left(m^2+m+1\right)=0$(Reduce to a complete square.)

$m^2\left(m^2+2·\frac{1}{2}·m+{\left(\frac{1}{2}\right)}^2+1-{\left(\frac{1}{2}\right)}^2\right)=0$(Square of sum reversed.)

$m^2\left({\left(m+\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}i\right)}^2\right)=0$(Difference of squares.)

$m^2\left(m+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\left(m+\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=0$

Therefore, roots of auxiliary equation are 0 with multiplicity $2,-\frac{1}{2}+\frac{\sqrt{3}}{2}{i}_{r}and-\frac{1}{2}-\frac{\sqrt{3}}{2}i$and .

Using the labels from Case III $\alpha =-\frac{1}{2},\beta =\frac{\sqrt{3}}{2}andm=0$

Since 0 has multiplicity of 2 we have $c_1{e}^{mx}+c_{2}x{e}^{mx}$as part of solution and because of non-real roots, we have $c_3{e}^{\alpha x}\cos \beta x+c_4{e}^{\alpha x}\sin \beta x$.

Therefore, the general solution is of the form:

$y=c_1{e}^{mx}+c_{2}x{e}^{mx}+c_3{e}^{\alpha x}\cos \beta x+c_4{e}^{\alpha x}\sin \beta x$(Substitute $\alpha ,\beta$ and m)

$y=c_1+c_{2}x+c_3{e}^{-x/2}\cos \frac{\sqrt{3}}{2}x+c_4{e}^{-x/2}\sin \frac{\sqrt{3}}{2}x$

Therefore, the solution is$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}\mathbf{+}{\mathit{c}}_{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathit{x}$