Q46E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

One way to define hyperbolic functions is by means of differential equations. Consider the equation $\mathit{y}\text{'}\text{'}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$. The hyperbolic cosine, $\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$, is defined as the solution of this equation subject to the initial values: $\mathit{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$ and $\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. The hyperbolic sine, $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$, is defined as the solution of this equation subject to the initial values: $\mathit{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$ and $\mathit{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

 

1. Solve these initial value problems to derive explicit formulas for $\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$ and $\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$. Also, show that $\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$ and $\frac{\mathit{d}}{\mathit{d}\mathit{t}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}$.
2. Prove that a general solution of the equation $\mathit{y}\text{'}\text{'}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$ is given by $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{t}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{t}$.
3. Suppose a,b and c are given constants for which $\mathit{a}{\mathit{r}}^{\mathbf{2}}\mathbf{+}\mathit{b}\mathit{r}\mathbf{+}\mathit{c}\mathbf{=}\mathbf{0}$ has two distinct real roots. If the two roots are expressed in the form $\mathit{\alpha }\mathbf{-}\mathit{\beta }$ and $\mathit{\alpha }\mathbf{+}\mathit{\beta }$, show that a general solution of the equation $\mathit{a}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{b}\mathit{y}\text{'}\mathbf{+}\mathit{c}\mathit{y}\mathbf{=}\mathbf{0}$ is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathit{\alpha }\mathit{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathbf{\left(}\mathit{\beta }\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathit{\alpha }\mathit{t}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{\left(}\mathit{\beta }\mathit{t}\mathbf{\right)}$.
4. Use the result of the part (c) to solve the initial value problem: $\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\text{'}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{17}\mathbf{/}\mathbf{2}$.

Step-by-Step Solution

Answer

1. One gets $cosh\left(t\right)=\frac{1}{2}{e}^t+\frac{1}{2}{e}^{-t}$ and $sinh\left(t\right)=\frac{1}{2}{e}^t-\frac{1}{2}{e}^{-t}$ So the derivatives produce each other $\frac{d}{dx}cosh\left(t\right)=\frac{d}{dx}\left(\frac{e^t+e^{-t}}{2}\right)=\frac{e^t-e^{-t}}{2}=sinh\left(t\right)$ and $\frac{d}{dx}sinh\left(t\right)=\frac{d}{dx}\left(\frac{e^t-e^{-t}}{2}\right)=\frac{e^t+e^{-t}}{2}=cosh\left(t\right)$.
2. If $y=c_{1}cosht+c_{2}sinht$ we get $y\text{'}\text{'}-y=0$.
3. One gets the general solution of $ay\text{'}\text{'}+by\text{'}+cy=0$ is $y=C_1{e}^{\alpha t}cosh\left(\beta t\right)+C_2{e}^{\alpha t}sinh\left(\beta t\right)$.
4. By using the part (c), we get $y_G=2e^{-\frac{t}{2}}cosh\left(-\frac{5}{2}t\right)+3e^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)$.

Step 1: Introduction

To solve for the explicit formulae, one considers the given and its initial values: $y\text{'}\text{'}-y=0$ where the initial values for $cosh\left(t\right)$ are $y\left(0\right)=1$ and $y\text{'}\left(0\right)=0$, the initial values for $sinh\left(t\right)$ are $y\left(0\right)=0$ and $y\text{'}\left(0\right)=1$.

To solve, first obtain the characteristic equation:

$$\begin{gathered} r^2-1=0 \\ r=\pm 1 \end{gathered}$$

 

Thus, one has the general solution $y_g=C_1{e}^t+C_2{e}^{-t}$ with its derivative $y{\text{'}}_g^{}=C_1{e}^t-C_2{e}^{-t}$.

One will use this equation to generate the system needed to solve for the specific cases of $sinh$ and $cosh$.

Step 2: Find the value of c o s h ( t )

First, one obtains

$$\begin{gathered} 1=C_1+C_2 \\ {C}_1=1-C_2 \\ 0=C_1-C_2 \end{gathered}$$ 

 

By substitution $0=1-2C_2$

Therefore, $C_2=\frac{1}{2}$. So $C_1=\frac{1}{2}$

Hence, $cosh\left(t\right)=\frac{1}{2}{e}^t+\frac{1}{2}{e}^{-t}$ or $\frac{e^t+e^{-t}}{2}$

Step 3: Find the value of sinh ( t )

Now for

 $$\begin{gathered} 0=C_1+C_2 \\ {C}_1=-C_2 \\ 1=C_1-C_2 \end{gathered}$$

By substitution $1=-2C_2$

So, $C_2=-\frac{1}{2}$ so $C_2=-\frac{1}{2}$

 

Thereafter, $sinh\left(t\right)=\frac{1}{2}{e}^t-\frac{1}{2}{e}^{-t}$or $\frac{{\mathrm{e}}^{\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}}{2}$

 

Now to prove the derivatives produce each other 

$\frac{d}{dx}cosh\left(t\right)=\frac{d}{dx}\left(\frac{e^t+e^{-t}}{2}\right)=\frac{e^t-e^{-t}}{2}=sinh\left(t\right)$ 

And

$\frac{d}{dx}sinh\left(t\right)=\frac{d}{dx}\left(\frac{e^t-e^{-t}}{2}\right)=\frac{e^t+e^{-t}}{2}=cosh\left(t\right)$

Step 4: Differentiate the value of y

To prove that $y=C_{1}cosh\left(t\right)+C_{2}sinh\left(t\right)$ is a general solution to $y\text{'}\text{'}-y=0$.

 

Consider:

$$\begin{gathered} y=C_{1}cosh\left(t\right)+C_{2}sinh\left(t\right) \\ y\text{'}=C_{1}sinh\left(t\right)+C_{2}cosh\left(t\right) \\ y\text{'}\text{'}=C_{1}cosh\left(t\right)+C_{2}sinh\left(t\right) \end{gathered}$$ 

 

By substitution $C_{1}cosh\left(t\right)+C_{2}sinh\left(t\right)-\left(C_{1}cosh\left(t\right)+C_{2}sinh\left(t\right)\right)\equiv 0$

Therefore, a general solution is $0\equiv 0$

Step 5: Checking roots

Now they want us to show that for a constant coefficient D.E. $a{r}^2+br+c=0$ which has 2 a real distinct root. We express these roots as $\alpha +\beta$ and $\alpha -\beta$ then we will have a general solution to $y=C_1{e}^{\alpha t}cosh\left(\beta t\right)+C_2{e}^{\alpha t}sinh\left(\beta t\right)$.

 

Note that we have conjugates for roots and that our general solution has the look of a complex solution. The big take away from this astonishing form of solution they are suggesting is that the complex solution will be tied up between the arguments of our function sinh and cosh and the powers of e.

 

$e^{\alpha t}$seems to escape out of the functions while we will attempt to factor $e^{\alpha t}$ out and use some tricks with sinh and cosh to obtain the desired result. Considering by $y=C_1{e}^{\alpha t}{e}^{\beta t}+C_2{e}^{\alpha t}{e}^{-\beta t}$ our roots to the auxiliary equation,

$$\begin{gathered} y=C_1{e}^{\alpha t}{e}^{\beta t}+C_2{e}^{\alpha t}{e}^{-\beta t} \\ y=e^{\alpha t}\left(C_1{e}^{\beta t}+C_2{e}^{-\beta t}\right) \end{gathered}$$

Step 6: Add and subtract c o s h ( β t ) , s i n h ( β t )

Now one will attempt to get $e^{\beta t}$ to relate to sinh and cosh

To get this relation we note that sinh and cosh are divided by 2. Therefore, we will add the two functions and subtract the 2 functions.

$$\begin{gathered} cosh\left(\beta t\right)+sinh\left(\beta t\right)=\frac{e^{\beta t}+e^{-\beta t}}{2}+\frac{e^{\beta t}-e^{-\beta t}}{2} \\ =\frac{2e^{\beta t}}{2} \\ =e^{\beta t} \end{gathered}$$ 

To know why $\beta$ goes to the power with t simply try out part a again only within the argument.

Now one tries $cosh\left(\beta t\right)-sinh\left(\beta t\right)$ because the negative will reverse the negative of sinh. 

$$\begin{gathered} cosh\left(\beta t\right)-sinh\left(\beta t\right)=\frac{e^{\beta t}+e^{-\beta t}}{2}-\frac{e^{\beta t}-e^{-\beta t}}{2} \\ =\frac{2e^{-\beta t}}{2} \\ =e^{-\beta t} \end{gathered}$$

 

Now that one has our identities found we can write the desired solution after a distribution $y=C_1{e}^{\alpha t}cosh\left(\beta t\right)+C_2{e}^{\alpha t}sinh\left(\beta t\right)$

Step 7: Finding the values of α , β

One uses our general solution to solve $y\text{'}\text{'}+y\text{'}-6y=0$ for $y\left(0\right)=2,y\text{'}\left(0\right)=-\frac{17}{2}$

The auxiliary equation gives: $r^2+r-6=0$

$\left(r+3\right)\left(r-2\right)=0$

 

Therefore, $r=-3$ and $r=2$ recall from the part c that the roots have the form $r=\alpha \pm \beta$ using this one can form a system. It doesn't matter which you set equal to what.

$$\begin{gathered} -3=\alpha + \\ \alpha =-3-\beta \\ 2=\alpha -\beta \end{gathered}$$ 

 

By substitution $2=-3-2\beta$

                      $-\frac{5}{2}=\beta$

So,  

$2=\alpha +\frac{5}{2}$             

Therefore, $\alpha =-\frac{1}{2}$

Step 8: Finding the values of C 1 , C 2

Our general solution is: $y_G=C_1{e}^{-\frac{t}{2}}cosh\left(-\frac{5}{2}t\right)+C_2{e}^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)$

$y{\text{'}}_G^{}=-\frac{1}{2}{C}_1{e}^{-\frac{t}{2}}cosh\left(-\frac{5}{2}t\right)-\frac{5}{2}{C}_1{e}^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)+\frac{1}{2}{C}_2{e}^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)-\frac{5}{2}{C}_2{e}^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)$ 

 

By our initial conditions: $2=C_1$

$$\begin{gathered} -\frac{17}{2}=-\frac{1}{2}{C}_1-\frac{5}{2}{C}_2 \\ -\frac{17}{2}=-\frac{1}{2}\left(2\right)-\frac{5}{2}{C}_2 \\ 3=C_2 \end{gathered}$$ 

 

Therefore, $y_G=2e^{-\frac{t}{2}}cosh\left(-\frac{5}{2}t\right)+3e^{-\frac{t}{2}}sinh\left(-\frac{5}{2}t\right)$.