Question

unit 7 practice packet date: 5. consider the differential equation \\( \frac{dy}{dx} = 6 - 2y \\). let \\( y = f(x) \\) be the particular solution to the differential equation with the initial condition \\( f(0) = 4 \\). a. write an equation for the line tangent to the graph of \\( y = f(x) \\) at \\( x = 0 \\). use the tangent line to approximate \\( f(0.6) \\) b. find the value of \\( \frac{d^2y}{dx^2} \\) at the point \\( (0, 4) \\). is the graph of \\( y = f(x) \\) concave up or concave down at the point \\( (0, 4) \\)? give a reason for the answer. c. find \\( y = f(x) \\), the particular solution to the differential equation with the initial condition \\( f(0) = 4 \\). d. for the particular solution \\( y = f(x) \\) found in part c, find \\( \lim_{x \to \infty} f(x) \\)

Explanation:

Part (a)

Step1: Find the slope at \( x = 0 \)

The differential equation is \( \frac{dy}{dx}=6 - 2y \). At \( x = 0 \), \( y = 4 \) (from \( f(0)=4 \)). Substitute \( y = 4 \) into \( \frac{dy}{dx} \):
\( \frac{dy}{dx}\bigg|_{x = 0}=6-2(4)=6 - 8=-2 \).

Step2: Write the tangent line equation

Using the point - slope form \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(0,4) \) and \( m=-2 \):
\( y - 4=-2(x - 0) \), which simplifies to \( y=-2x + 4 \).

Step3: Approximate \( f(0.6) \)

Substitute \( x = 0.6 \) into the tangent line equation \( y=-2x + 4 \):
\( y=-2(0.6)+4=-1.2 + 4 = 2.8 \).

Part (b)

Step1: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^{2}y}{dx^{2}} \)

Given \( \frac{dy}{dx}=6 - 2y \), differentiate both sides with respect to \( x \) using the chain rule. The derivative of \( 6 \) with respect to \( x \) is \( 0 \), and the derivative of \( - 2y \) with respect to \( x \) is \( -2\frac{dy}{dx} \). So, \( \frac{d^{2}y}{dx^{2}}=-2\frac{dy}{dx} \).

Step2: Substitute \( x = 0,y = 4 \) and \( \frac{dy}{dx}\bigg|_{(0,4)}=-2 \)

Substitute \( \frac{dy}{dx}=-2 \) (from part (a)) into the second - derivative formula:
\( \frac{d^{2}y}{dx^{2}}\bigg|_{(0,4)}=-2(-2)=4 \).

Step3: Determine concavity

Since \( \frac{d^{2}y}{dx^{2}}\bigg|_{(0,4)} = 4>0 \), the graph of \( y = f(x) \) is concave up at \( (0,4) \) (because the second derivative is positive at that point).

Part (c)

Step1: Recognize the type of differential equation

The differential equation \( \frac{dy}{dx}=6 - 2y \) is a separable differential equation. Rewrite it as:
\( \frac{dy}{6 - 2y}=dx \).

Step2: Integrate both sides

Integrate the left - hand side with respect to \( y \) and the right - hand side with respect to \( x \).
For the left - hand side, let \( u = 6-2y \), then \( du=-2dy \) or \( dy=-\frac{1}{2}du \).
\( \int\frac{dy}{6 - 2y}=\int-\frac{1}{2}\frac{du}{u}=-\frac{1}{2}\ln|u|+C_1=-\frac{1}{2}\ln|6 - 2y|+C_1 \).
The right - hand side: \( \int dx=x + C_2 \).
Combining the constants \( C = C_2 - C_1 \), we have \( -\frac{1}{2}\ln|6 - 2y|=x + C \).
Multiply both sides by \( - 2 \): \( \ln|6 - 2y|=-2x + C' \) (where \( C'=-2C \)).
Exponentiate both sides: \( |6 - 2y|=e^{-2x + C'}=e^{C'}e^{-2x} \). Let \( A = e^{C'} \) ( \( A>0 \) ), so \( 6 - 2y=\pm A e^{-2x} \). We can write it as \( 6 - 2y=Ke^{-2x} \) (where \( K=\pm A \), \( K
eq0 \) initially, but we will find \( K \) using the initial condition).

Step3: Use the initial condition \( f(0)=4 \)

When \( x = 0 \), \( y = 4 \). Substitute into \( 6 - 2y=Ke^{-2x} \):
\( 6-2(4)=K e^{0}\Rightarrow6 - 8 = K\Rightarrow K=-2 \).
So, \( 6 - 2y=-2e^{-2x} \).
Solve for \( y \):
\( - 2y=-2e^{-2x}-6\Rightarrow y = e^{-2x}+3 \).

Part (d)

Answer:

Step1: Find the limit of \( y = e^{-2x}+3 \) as \( x

ightarrow\infty \)
We know that for the exponential function \( y = e^{-2x}=\frac{1}{e^{2x}} \), as \( x
ightarrow\infty \), \( e^{2x}
ightarrow\infty \), so \( \frac{1}{e^{2x}}
ightarrow0 \).
Then, \( \lim_{x
ightarrow\infty}f(x)=\lim_{x
ightarrow\infty}(e^{-2x}+3)=\lim_{x
ightarrow\infty}e^{-2x}+\lim_{x
ightarrow\infty}3 \).
Since \( \lim_{x
ightarrow\infty}e^{-2x}=0 \) and \( \lim_{x
ightarrow\infty}3 = 3 \), we have \( \lim_{x
ightarrow\infty}f(x)=3 \).

Final Answers

a. The equation of the tangent line is \( y=-2x + 4 \) and \( f(0.6)\approx2.8 \).
b. \( \frac{d^{2}y}{dx^{2}}\bigg|_{(0,4)} = 4 \), and the graph is concave up at \( (0,4) \) because the second derivative is positive.
c. \( y = f(x)=e^{-2x}+3 \).
d. \( \lim_{x
ightarrow\infty}f(x)=3 \).