91Ó°ÊÓ

Given is$\mathrm{Ó\neg }=1,b=\frac{13}{5},1,\frac{5}{13}$

The conditions for these cases are,

-For the overdamped case-, $b^2>{\mathrm{Ó\neg }}^2$and the solution for it is,$y=A{e}^{-\mathrm{λ}t}+B{e}^{\mathrm{μ}t}$ where,$\mathit{λ}\mathbf{=}\mathit{b}\mathbf{+}\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}{\mathbf{Ó\neg }}^{\mathbf{2}}}$ and.$\mathrm{μ}=b-\sqrt{b^2-{\mathrm{Ó\neg }}^2}$

-For the critically damped case- the condition is$b^2={\mathrm{Ó\neg }}^2$ , and the solution is $y=(A+Bt)e^{-bt}$.

-For the underdamped (oscillatory)- case the condition is$b^2<{\mathrm{Ó\neg }}^2$ , and the solution is$y=e^{-bt}(Asin\mathrm{β}t+Bcos\mathrm{β}t)$ .

For the overdamped case-we will choose$\mathrm{Ó\neg }=1$ and.$b=13/5$ Therefore, the value of$\mathrm{λ}=5$ & $\mathrm{μ}=1/5$.

(a)

Find the first derivative to apply the initial condition to find the value of the constants in the solution,

$y^{\text{'}}=−\mathrm{λ}A{e}^{−\mathrm{λ}t}−\mathrm{μ}B{e}^{−\mathrm{μ}t}$

Now, we apply the initial conditions,

$\begin{array}{c}y\left(0\right)=1\\ =A+B\\ A=1−B\mathrm{............}\left(1\right)\end{array}$

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=0\\ =−\mathrm{λ}A−\mathrm{μ}B\mathrm{.......}\left(2\right)\end{array}$

Substitute $A$into eq.(2), so we get

$\begin{array}{l}−\mathrm{λ}+\mathrm{λ}B−\mathrm{μ}B\\ =0B(\mathrm{λ}−\mathrm{μ})\\ =\mathrm{λ}B\\ =\frac{\mathrm{λ}}{\mathrm{λ}−\mathrm{μ}}\end{array}$

Solve further

$\begin{array}{l}=\frac{5}{5−1/5}\\ =\frac{1}{24}\end{array}$

therefore, using eq. (1)

$A=\frac{23}{24}$

So, the solution is

$y=\frac{23}{24}{e}^{−5t}+\frac{1}{24}{e}^{−t/5}$

Graph is

For the critically damped case we choose, $\mathrm{Ó\neg }=b=1$and we derive the solution to get

$y^{\text{'}}=B{e}^{−bt}−b(A+Bt)e^{−bt}$

Now, applying the boundary conditions, so we get

$\begin{array}{c}y\left(0\right)=1\\ =A\end{array}$

Now,

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=0\\ =B−bA\\ B=1\end{array}$

So, the solution is

$y=(1+t)e^{−t}$

Graph is

For the oscillatory case we choose, $\mathrm{Ó\neg }=1$and $b=5/13$, so the value of $\mathrm{β}=12/13$, and the first derivative of the general solution is

$y^{\text{'}}=−b{e}^{−bt}(A\sin \mathrm{β}t+B\cos \mathrm{β}t)+e^{−bt}\mathrm{β}(A\cos \mathrm{β}t−\sin \mathrm{β}t)$

Now, apply the initial conditions,

$\begin{array}{l}y\left(0\right)=1\\ =B{y}^{\text{'}}\left(0\right)\\ =0\\ =−bB+\mathrm{β}AA\end{array}$

Solve further

$\begin{array}{l}=\frac{bB}{\mathrm{β}}\\ =\frac{5}{12}\end{array}$

So, the solution now is

$y=e^{−5t/13}(\frac{5}{12}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

Graph is

Combine the three, to get

This time the initial conditions are$y\left(0\right)=1$and $y^{\text{'}}\left(0\right)=1$. The constants will have the value as part). For the overdamped case (i.e. $\mathrm{Ó\neg }=1$,, and$b=13/5$) and have

$\begin{array}{c}y\left(0\right)=1\\ =A+B\\ =1\text{ }\\ B=1−A\end{array}$

Now,

$\begin{array}{c}{y}^{\text{'}}\left(0\right)=1\\ =−5A−\frac{1}{5}\text{ }\\ =−5A−\frac{1}{5}+\frac{1}{5}\end{array}$

Therefore,

$\begin{array}{c}A=1\\ A=−\frac{1}{4},\text{and}B=\frac{5}{4}\end{array}$

So, the solution is

$y=−\frac{1}{4}{e}^{−5t}+\frac{5}{4}{e}^{−t/5}$

Graph is

For the critical damped case i.e $(\mathrm{Ó\neg }=b=1)$.,we have

$\begin{array}{c}y\left(0\right)=1\\ =A\\ y^{\text{'}}\left(0\right)=1\\ =B−1\text{ }\\ B=2\end{array}$

So, the solution becomes

$y=(1+2t)e^{−t}$

Graph is

For the oscillatory case (i.e.,$\mathrm{Ó\neg }=1,and$) have

$\begin{array}{c}y\left(0\right)=1\\ =B{y}^{\text{'}}\\ =1\\ =−\frac{5}{13}+\frac{12}{13}A\end{array}$

Solve further

$A=\frac{6}{13}$

So, the solution is

$y=e^{−5t/13}(\frac{6}{13}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

The graph is

And when combine the three graphs to get.

This time the initial conditions are $y\left(0\right)=1$and$y^{\text{'}}\left(0\right)=−1$the constants will have the value as part (a). For the overdamped case (i.e $\mathrm{Ó\neg }=1$.,, and$b=13/5$) have

$\begin{array}{c}y\left(0\right)=1=A+B\text{ }\\ A=1−B\\ y^{\text{'}}\left(0\right)=−1\\ =−5+5B−\frac{1}{5}B\end{array}$

So, the solution is

$∴A=−\frac{76}{24},\text{and}B=\frac{100}{24}$

Graph is

For the critically damped case (i.e. $\mathrm{Ó\neg }=b=1$, ) have

$\begin{array}{c}y\left(0\right)=1=A\\ y^{\text{'}}\left(0\right)=−1\\ =B−1\text{ }\\ \text{ }B=0\end{array}$

So, the solution is

$y=e^{−t}$

Graph is

For the oscillatory case (i.e.,$\mathrm{Ó\neg }=1,\text{ }\&b=5/13$) and have

$\begin{array}{c}y\left(0\right)=1=B\\ y^{\text{'}}\left(0\right)=−1\\ =−\frac{5}{13}+\frac{12}{13}A\text{ }\\ \text{ }A=−\frac{2}{3}\end{array}$

So, the solution is

$y=e^{−5t/13}(−\frac{2}{3}\sin \frac{12}{13}t+\cos \frac{12}{13}t)$

Graph is

Combine the graph three, and get