Chapter 6, Problem 66RE

$\text{(a)}$

To determine

To find: The equation for the line tangent to the graph of $f$ at the point $(1,0)$ .

Answer to Problem 66RE

The answer:

  $\begin{array}{c}y=-3x+3\\ f(1.2)\approx -0.6\end{array}$

Explanation of Solution

Given information:

The differential equation is $dy/dx=e^y\left(3x^2-6x\right)$ .

  $y=f(x)$ , the particular solution to the differential equation passes through $(1,0)$ .

Calculation:

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

  $\begin{array}{c}\frac{dy}{dx}=e^y\left(3x^2-6x\right)\\ =e^0\left(3{(1)}^2-6(1)\right)\\ =-3\\ \to m=-3\end{array}$

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

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ \to y-0=-3(x-1)\\ \to y=-3x+3\end{array}$

Now,

Put $x=1.2$ to approximate $f(1.2)$:

Then,

  $\begin{array}{c}y=-3x+3\\ =-3(1.2)+3\\ =-0.6\\ \to f(1.2)\approx -0.6\end{array}$

Therefore, the required equation for the line tangent to the graph of $f$ at the point $(1,0)$ .

  $\begin{array}{c}y=-3x+3\\ f(1.2)\approx -0.6\end{array}$

$\text{(b)}$

To determine

To find: The particular solution to the differential equation that passes through $(1,0)$ .

Answer to Problem 66RE

The answer: $y=-\ln \left(-x^3+3x^2-1\right)$

Explanation of Solution

Given information:

The differential equation is $dy/dx=e^y\left(3x^2-6x\right)$ .

  $y=f(x)$ , the particular solution to the differential equation passes through $(1,0)$ .

Calculation:

  $\begin{array}{c}\frac{dy}{dx}=e^y\left(3x^2-6x\right)\\ \to \frac{dy}{e^y}=\left(3x^2-6x\right)dx\\ \to e^{-y}dy=\left(3x^2-6x\right)dx\\ \to -e^{-y}=x^3-3x^2+C\\ =3x^2-x^3+C_1\\ -y=\ln \left(3x^2-x^3+C_1\right)\\ y=-\ln \left(3x^2-x^3+C_1\right)\end{array}$

Plug in the initial condition, which is point $(1,0)$ to find $C_1$ :

  $\begin{array}{c}0=-\ln \left(3{(1)}^2-{(1)}^3+C_1\right)\\ =\ln \left(2+C_1\right)\\ \to e^0=2+C_1\\ \to 1=2+C_1\\ \to C_1=-1\end{array}$

Therefore, the required particular solution to the differential equation that passes through $(1,0)$ is:

  $y=-\ln \left(-x^3+3x^2-1\right)$