Find a general solution to the following higher-order equations.

(a)$\mathit{y}\text{'}\text{'}\text{'}\mathbf{-}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{y}\text{'}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0}$

(b)$\mathit{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathit{y}\text{'}\text{'}\mathbf{+}\mathbf{5}\mathit{y}\text{'}\mathbf{-}\mathbf{26}\mathit{y}\mathbf{=}\mathbf{0}$

(c)${\mathit{y}}^{iv}\mathbf{+}\mathbf{13}\mathit{y}\text{'}\text{'}\mathbf{+}\mathbf{36}\mathit{y}\mathbf{=}\mathbf{0}$

(a)

The auxiliary equation is $r^3-r^2+r+3=0$.

It is not difficult to see that one root of this equation is $r=-1$and dividing the previous equation with $r+1$we get that $\left(r^2-r^2+r+3\right):(r+1)=r^2-2r+3$if and only if $r^3-r^2+r+3=(r+1)\left(r^2-2r+3\right)$

Now we need to determine the general solution of the given differential equation:

$y\left(t\right)=c_1{e}^{r_{1}t}+c_2{e}^{\alpha t}cos\beta t+c_3{e}^{\alpha t}sin\beta t$

.

Where $r_{2,3}=a\pm ib$. We see that $\alpha =1$and $\beta =\sqrt{2}$, so the general solution is $y(t)=c_1{e}^{-t}+c_2{e}^{t}cos\sqrt{2}t+c_3{e}^{t}sin\sqrt{2}t$.

(b)

The auxiliary equation is$r^3+2r^2+5r-26=0$. It is not difficult to see that one root of this equation is$r_1=2$ and dividing the previous equation with$r-2$ we get that $\left({\mathrm{r}}^3+2{\mathrm{r}}^2+5\mathrm{r}-26\right):(\mathrm{r}-2)={\mathrm{r}}^2+4\mathrm{r}+13$

If and only if $r^3+2r^2+5r-26=(r-2)\left(r^2+4r+13\right)$

Now we need to determine the general solution of the given differential equation:

$y\left(t\right)=c_1{e}^{r_{1}t}+c_2{e}^{\alpha t}cos\beta t+c_3{e}^{\alpha t}sin\beta t$where $r_{2,3}=\alpha \pm i\beta$.

We see that $\alpha =-2$and $\beta =3$, so the general solution is $y\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}cos3t+c_3{e}^{-2t}sin3t$.

(c)

The auxiliary equation is $r^4+13r^2+36=0$. To find the roots of this equation we will take substitution $t=r^2$

$$\begin{gathered} t^2+13t+36=0 \\ {t}_{1,2}=\frac{-13\pm \sqrt{169-144}}{2} \\ =\frac{-13\pm 5}{2} \\ {t}_1=-4, \\ {t}_2=-9 \end{gathered}$$

Now we can find all four roots of the auxiliary equation:

$$\begin{gathered} r^2=t_1\Rightarrow r^2=-4\Rightarrow r_{1,2}=\pm 2i \\ {r}^2=t_2\Rightarrow r^2=-9\Rightarrow r_{3,4}=\pm 3i \end{gathered}$$

Since all roots of the auxiliary equation are complex, we have that the general solution of this differential equation is:

$y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t+c_3{e}^{\gamma t}cos\delta t+c_4{e}^{\gamma t}sin\delta t$, where $r_{1,2}=\alpha \pm \beta i$and $r_{3,4}=\gamma \pm \delta i$.

We have that $\alpha =\gamma =0,\beta =2$and $\delta =3$, so the general solution of the given differential equation is:

$y\left(t\right)=c_{1}cos2t+c_{2}sin2t+c_{3}sin3t+c_{4}sin3t$.