Chapter 8.6, Problem 1P

(a)

To determine

To show: The solution is $y={\varphi }_1\left(t\right)=2-t$ by using the given data.

Explanation of Solution

Consider the initial value problem is $y\text{'}=t+y-3,y\left(0\right)=2$.

Therefore, the given solution is $y={\varphi }_1\left(t\right)=2-t$.

Differentiate the equation $y={\varphi }_1\left(t\right)=2-t$ with respect to t.

$\begin{array}{c}\frac{d}{dt}\left(y\right)=\frac{d}{dt}\left(2-t\right)\\ y\text{'}=-1\end{array}$

Substitute the value of the $y=2-t$ and $y\text{'}=-1$ into the initial problem to verify that it is a solution.

$\begin{array}{c}y\text{'}=t+y-3\\ -1=t+2-t-3\\ =0\end{array}$

Hence, it is a solution of the initial problem. Lastly, verify that it satisfies the initial condition $y\left(0\right)=2$.

Substitute the value of $t=0$ in $y={\varphi }_1\left(t\right)=2-t$.

$\begin{array}{l}{\varphi }_1\left(t\right)=2-0\\ {\varphi }_1\left(t\right)=2\end{array}$

Hence, the value of ${\varphi }_1\left(t\right)=2$ satisfy the initial condition.

(b)

To determine

The solution $y={\varphi }_2\left(t\right)$ and compare the difference ${\varphi }_2\left(t\right)-{\varphi }_1\left(t\right)$ at $t=1$ and obtain the value as $t\to \infty$ by using the given data.

Answer to Problem 1P

The solution $y={\varphi }_2\left(t\right)$ is $\underset{\_}{{\varphi }_2\left(t\right)=0.001e^t+2-t}$ and the difference ${\varphi }_2\left(t\right)-{\varphi }_1\left(t\right)$ is $\underset{\_}{\underset{t\to \infty }{\lim }{\varphi }_2\left(1\right)-{\varphi }_1\left(1\right)=\underset{t\to \infty }{\lim }0.001e^t=\infty }$.

Explanation of Solution

Consider the initial value problem $y\text{'}=t+y-3,y\left(0\right)=2$.

First, to calculate the corresponding homogeneous equation $y\text{'}=y$ since the required solution is $y_h=c{e}^t$.

The particular solution is of the form $y_p=A+Bt$.

Since, it must be a solution of the initial problem.

In order to calculate the value of $A\text{ and }B$, substitute it into the $y\text{'}=t+y-3$.

$\begin{array}{l}{y}_p^{\text{'}}=t+A+Bt-3\\ A=2,B=-1\end{array}$

The solution of the initial ODE is given as $y=y_h+y_p$, thus the general solution is $y\left(t\right)=c{e}^t+2-t$ but the initial condition is given as $y\left(0\right)=2.001$.

Substitute the value $t=0$ in $y\left(t\right)=c{e}^t+2-t$.

$\begin{array}{l}y\left(0\right)=c{e}^0+2-0\\ 2.001=c+2\\ c=0.001\end{array}$

Substitute the value of $c=0.001$ in $y\left(t\right)=c{e}^t+2-t$.

$y\left(t\right)={\varphi }_2\left(t\right)=0.001e^t+2-t$

Therefore, the difference at $t=1$ is $\underset{t\to \infty }{\lim }{\varphi }_2\left(1\right)-{\varphi }_1\left(1\right)=\underset{t\to \infty }{\lim }0.001e^t=\infty$

Hence, the solution $y={\varphi }_2\left(t\right)$ is $\underset{\_}{{\varphi }_2\left(t\right)=0.001e^t+2-t}$ and the difference ${\varphi }_2\left(t\right)-{\varphi }_1\left(t\right)$ is $\underset{\_}{\underset{t\to \infty }{\lim }{\varphi }_2\left(1\right)-{\varphi }_1\left(1\right)=\underset{t\to \infty }{\lim }0.001e^t=\infty }$.