
Question content area top
Part 1
Show that variant phi left parenthesis x right parenthesis equals Upper C e Superscript negative 4 x Baseline plus 3
 
is a solution to StartFraction dy Over dx EndFraction plus 4 y equals 12
 
for any choice of the constant C.​ Thus, Upper C e Superscript negative 4 x Baseline plus 3
 
is a​ one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.
Question content area bottom
Part 1
Substitute variant phi
​(x) 
for y and find the derivative to substitute for StartFraction dy Over dx EndFraction
.
StartFraction dy Over dx EndFraction
plus 4 y
equals
12
enter your response here
plus 4left parenthesis Upper C e Superscript negative 4 x Baseline plus 3 right parenthesis
equals
12

Let's begin by verifying that \( \varphi(x) = C e^{-4x} + 3 \) is a solution to the differential equation.

### Step 1: Substitute \( \varphi(x) \) for \( y \)
We are given the differential equation:
\[
\frac{dy}{dx} + 4y = 12
\]
Substitute \( \varphi(x) = C e^{-4x} + 3 \) for \( y \):
\[
\frac{d}{dx} ( C e^{-4x} + 3 ) + 4(C e^{-4x} + 3) = 12
\]

### Step 2: Find the derivative \( \frac{dy}{dx} \)
To find the derivative of \( \varphi(x) = C e^{-4x} + 3 \), apply the chain rule:
\[
\frac{d}{dx} \left( C e^{-4x} + 3 \right) = C \cdot (-4) e^{-4x} = -4C e^{-4x}
\]

### Step 3: Substitute \( \frac{dy}{dx} \) into the equation
Now, substitute \( \frac{dy}{dx} = -4C e^{-4x} \) into the original differential equation:
\[
-4C e^{-4x} + 4(C e^{-4x} + 3) = 12
\]

### Step 4: Simplify the equation
Expand and simplify the terms:
\[
-4C e^{-4x} + 4C e^{-4x} + 12 = 12
\]
The terms \( -4C e^{-4x} \) and \( 4C e^{-4x} \) cancel out:
\[
12 = 12
\]

This is true for any value of \( C \), which shows that \( \varphi(x) = C e^{-4x} + 3 \) is indeed a solution to the differential equation.

### Conclusion
Since the equation holds true for any constant \( C \), the function \( \varphi(x) = C e^{-4x} + 3 \) represents a one-parameter family of solutions to the differential equation \( \frac{dy}{dx} + 4y = 12 \).

### Graphing
To visualize the solution curves, you can graph several instances of \( \varphi(x) = C e^{-4x} + 3 \) for different values of \( C \) on the same coordinate axes. For example, try \( C = 1, C = 2, C = -1 \), and so on. All curves will approach \( y = 3 \) as \( x \to \infty \), but their behavior for small values of \( x \) will differ depending on \( C \).

---

Would you like to see the graph for this family of solutions? Let me know if you need further details!

### Related Questions:
1. How do the solution curves change as the constant \( C \) increases or decreases?
2. What is the behavior of the solutions as \( x \to \infty \) for different values of \( C \)?
3. How would you solve the differential equation using an integrating factor method?
4. Could you describe the physical interpretation of the differential equation \( \frac{dy}{dx} + 4y = 12 \)?
5. How do you find the general solution to a first-order linear differential equation?

### Tip:
When solving a differential equation, always verify your solution by substituting it back into the original equation to ensure that all terms cancel out properly.

Show that ϕ(x) = Ce^(-4x) + 3 is a solution to dy/dx + 4y = 12 for any choice of the constant C. Thus, Ce^(-4x) + 3 is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.

Explore the step-by-step solution to the differential equation dy/dx + 4y = 12, showing that ϕ(x) = Ce^(-4x) + 3 is a one-parameter family of solutions. This problem involves first-order linear equations and is ideal for undergraduate students.

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