Chapter 6, Problem 65RE

$\text{(a)}$

To determine

To sketch: The solution curve through the point $(0,1)$ .

Answer to Problem 65RE

The graph is shown:

  

Explanation of Solution

Given information:

The slope filed is:

  

Calculation:

Take at point $(0,1)$ , draw one line to the right and one line to the left, each following the slope field.

  

$\text{(b)}$

To determine

To find: The equation for the line tangent to the solution given in part $\text{(a)}$ at the point $(0,1)$ .

Answer to Problem 65RE

The answer:

  $\begin{array}{c}y=2x+1\\ f(0.2)\approx 1.4\end{array}$

Explanation of Solution

Given information:

The slope filed is:

  

Calculation:

The slope of the line equals $dy/dx$ find the slope of the tangent line, plug $(0,1)$ into $dy/dx$ .

  $\begin{array}{c}\frac{dy}{dx}=(3-y)\cos x\\ =(3-1)\cos 0\\ =2\\ \to m=2\end{array}$

Now use the slope, point $(0,1)$ , and the point slope formula to find the equation for the tangent line:

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ \to y-1=2(x-0)\\ \to y=2x+1\end{array}$

Now,

Put $x=0.2$ to approximate $f(0.2)$ .

Then,

  $\begin{array}{c}y=2x+1\\ =2(0.2)+1\\ =1.4\\ \to f(0.2)\approx 1.4\end{array}$

Therefore, the required equation for the line tangent to the solution given in part $\text{(a)}$ at the point $(0,1)$ is:

  $\begin{array}{c}y=2x+1\\ f(0.2)\approx 1.4\end{array}$

$(c)$

To determine

To find: The particular solution to the differential equation with initial condition $f(0)=1$ .

Answer to Problem 65RE

The answer: $y=3-2e^{-\sin x}$

Explanation of Solution

Given information:

The slope filed is:

  

Calculation:

  $\begin{array}{c}\frac{dy}{dx}=(3-y)\cos x\\ \to \frac{dy}{3-y}=\cos xdx\\ \to -\ln (3-y)=\sin x+C\\ =-\sin x+C_1\\ 3-y=e^{-\sin x+C_1}\\ =C_2{e}^{-\sin x}\\ \to y=3-C_2{e}^{-\sin x}\end{array}$

Put the initial condition

  $f(0)=1$ point $(0,1)$

To find $C_2$:

Then,

  $\begin{array}{c}1=3-C_2{e}^{-\sin 0}\\ =3-C_2\\ \to C_2=2\end{array}$

Therefore, the required particular solution to the differential equation with initial condition $f(0)=1$ is:

  $y=3-2e^{-\sin x}$