Q29E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

Step-by-Step Solution

Answer

(a) The general solution of the given initial value $y\text{'}\text{'}\text{'}-y\text{'}\text{'}+y\text{'}+3y=0$ is: $y\left(t\right)=c_1{e}^{-t}+c_2{e}^{t}cos\sqrt{2}t+c_3{e}^{t}sin\sqrt{2}t$

  

(b) The general solution of the given initial value $y\text{'}\text{'}\text{'}+2y\text{'}\text{'}+5y\text{'}-26y=0$ is: $y\left(t\right)=c_1{e}^{-t}+c_2{e}^{t}cos\sqrt{2}t+c_3{e}^{t}sin\sqrt{2}t$

  

(c) The general solution of the given initial value $y\text{'}\text{'}\text{'}+13y\text{'}\text{'}+36y=0$ is: $y\left(t\right)=c_{1}cos2t+c_{2}sin2t+c_{3}sin3t+c_{4}sin3t$

Step 1: Finding the roots.

 (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 $y(t)=c_1{e}^{-t}+c_2{e}^{t}cos\sqrt{2}t+c_3{e}^{t}sin\sqrt{2}t$.

Step 2: Finding the roots

(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 $y\left(t\right)=c_1{e}^{2t}+c_2{e}^{-2t}cos3t+c_3{e}^{-2t}sin3t$.

Step 3: Finding the roots

(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:

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

 

Step 4: Finding the general solution

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:

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