Problem 52 - Calculus - AP Edition

Question

Find a function \(f\) such that \(y=f(x)\) satisfies \(d y / d x=e^{y}\) with \(y=-\ln 2\) when \(x=0\)

Step-by-Step Solution

Answer

The function is \( y = -\ln(2 - x) \).

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{dy}{dx} = e^y \). We need to find the function \( y = f(x) \) that satisfies this equation as well as the initial condition \( y = -\ln 2 \) when \( x = 0 \).

Step 2: Separate Variables

To solve the equation, we'll separate variables by rewriting it as \( \frac{dy}{e^y} = dx \). This allows us to integrate both sides independently.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[ \int \frac{1}{e^y} \, dy = \int 1 \, dx \]The left side integrates to \( -e^{-y} \) and the right to \( x + C \), giving us the equation: \( -e^{-y} = x + C \).

Step 4: Solve for Integration Constant

Apply the initial condition \( y = -\ln 2 \) when \( x = 0 \) to find \( C \):\( -e^{-(-\ln 2)} = 0 + C \)\( -2 = C \).

Step 5: Form the General Solution

Substitute \( C = -2 \) back into the equation:\[ -e^{-y} = x - 2 \].

Step 6: Solve for y

Rearrange the equation to solve for \( y \):\[ e^{-y} = 2 - x \]Taking the natural logarithm on both sides:\[ -y = \ln(2 - x) \]Therefore,\[ y = -\ln(2 - x) \].

Key Concepts

Separation of Variables

In the context of differential equations, separation of variables is a powerful method used to solve first-order equations. The idea behind this technique is to rearrange the equation such that all terms involving one variable and its derivative are on one side, while the terms involving the other variable are on the opposite side.
This allows you to integrate each side concerning its particular variable.
For the differential equation \( \frac{dy}{dx} = e^y \), we start by separating variables to make it easier to handle. The separation looks like this:
\[\frac{1}{e^y} \, dy = dx\] Now, each side of this equation is ready to be integrated independently. Separation of variables is a crucial step because it makes the equation simpler to solve by focusing on one variable at a time. Once separated, we can proceed to integrate both sides.

Integration

Integration is a core concept used to solve differential equations, especially after separating variables. Integration lets us find a function whose derivative matches the original function given.
After separating variables in the exercise, we integrate each side concerning its variable:
\[ \int \frac{1}{e^y} \, dy = \int 1 \, dx \] This results in:

• Left side: the integration \( \int \frac{1}{e^y} \, dy \) becomes \( - e^{-y} \)
• Right side: the integration \( \int 1 \, dx \) results in \( x + C \), where \( C \) is the integration constant

By integrating both sides, we find a relationship between \( y \) and \( x \), which can then be adjusted using initial conditions to solve for the specific function that satisfies the differential equation.

Initial Conditions

Initial conditions play a critical role in solving differential equations, especially when determining the integration constant \( C \).
An initial condition is a given value for the function and its independent variable, ensuring the solution is unique and specific.
In this problem, the initial condition is \( y = -\ln 2 \) when \( x = 0 \). We use this to substitute into the equation:

• Plug \( y = -\ln 2 \) and \( x = 0 \) into the integrated equation: \(-e^{-(-\ln 2)} = 0 + C\)
• Simplifying, we find \(-2 = C\)

This value of \( C \) helps narrow down the general solution \(-e^{-y} = x + C\) to a specific function, giving the exact pattern that matches the initial conditions. The result makes our solution specific, fitting perfectly within the boundary set by real-world or theoretical problems.