Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, solve each differential equation. $$ y^{\prime}=e^{2 x}-3 y ; y=1 \text { when } x=0 $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{1}{5} e^{2x} + \frac{4}{5} e^{-3x} \).

Step 1: Identify the Differential Equation Type

The equation given is a first-order linear differential equation of the form \( y' + py = q \). Here, it is \( y' = e^{2x} - 3y \), which can be rewritten as \( y' + 3y = e^{2x} \). Since it is linear, we can use an integrating factor to solve it.

Step 2: Determine the Integrating Factor

The standard form of a first-order linear differential equation is \( y' + P(x) y = Q(x) \). In this case, \( P(x) = 3 \). The integrating factor \( \mu(x) \) is found using \( \mu(x) = e^{\int P(x) \, dx} = e^{\int 3 \, dx} = e^{3x} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiplying the entire differential equation by the integrating factor, we have: \( e^{3x} y' + 3 e^{3x} y = e^{3x} e^{2x} \). Which simplifies to: \( e^{3x} y' + 3 e^{3x} y = e^{5x} \).

Step 4: Recognize and Solve the Left Side as a Derivative

The left-hand side of the equation \( e^{3x} y' + 3 e^{3x} y \) can be written as the derivative of a product: \( \frac{d}{dx}(e^{3x} y) \). Thus, the equation becomes \( \frac{d}{dx}(e^{3x} y) = e^{5x} \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \). \( \int \frac{d}{dx}(e^{3x} y) \, dx = \int e^{5x} \, dx \)The left side integrates to \( e^{3x} y \), and the right side integrates to \( \frac{1}{5} e^{5x} + C \), where \( C \) is the constant of integration. So, we have: \( e^{3x} y = \frac{1}{5} e^{5x} + C \).

Step 6: Solve for \( y \)

Now, isolate \( y \):\[ y = \frac{1}{5} e^{2x} + Ce^{-3x} \].

Step 7: Use Initial Condition to Find Constant

Apply the initial condition \( y(0) = 1 \):\[ 1 = \frac{1}{5} e^{0} + Ce^{0} \]\[ 1 = \frac{1}{5} + C \]\[ C = 1 - \frac{1}{5} = \frac{4}{5} \].

Step 8: Write the Final Solution

Substitute the value of \( C \) back into the equation: \[ y = \frac{1}{5} e^{2x} + \frac{4}{5} e^{-3x} \].

Key Concepts

First-order linear differential equations

Differential equations describe how a function changes and can model many real-world phenomena. When dealing with first-order linear differential equations, we work with equations where the highest derivative is the first one. Typically, these take the form \( y' + P(x)y = Q(x) \). In these equations:

• \( y' \) represents the derivative of \( y \) with respect to \( x \).
• \( P(x) \) and \( Q(x) \) are functions of the independent variable \( x \).

They are termed 'linear' because both the dependent variable (\( y \)) and its derivative (\( y' \)) appear to the first power, and there's no product of \( y \) terms (like \( y^2 \)). In this exercise, the given equation fits this mold and can be solved using specific techniques.

Integrating factor

An integral part of solving first-order linear differential equations is finding the integrating factor. This factor simplifies the equation into one that can be easily integrated. The integrating factor, \( \mu(x) \), is derived from \( P(x) \):

• First, find \( P(x) \) from the equation \( y' + P(x)y = Q(x) \).
• Calculate \( \mu(x) = e^{\int P(x) \, dx} \).

For the equation at hand, \( P(x) = 3 \), leading to the integrating factor \( \mu(x) = e^{3x} \). The use of \( \mu(x) \) helps convert the left-hand side of the differential equation into a single derivative, making the equation easier to integrate and solve.

Initial conditions

Initial conditions provide specific information regarding the value of the solution at a certain point. They play a crucial role in determining the particular solution of a differential equation. Given an initial condition like \( y = 1 \) when \( x = 0 \), we:

• Substitute these values into the general solution to find the constant of integration (\( C \)).
• Ensure the solution meets these initial specifications, resulting in a unique solution.

This process ensures that despite many potential solutions the differential equation might have, the initial conditions help narrow it down to one specific solution that fits the given criteria.

Calculus solutions

Solving differential equations often relies on calculus, specifically on integration and differentiation. Once the differential equation is set up with its integrating factor, calculus is used to:

• Recognize the left-hand side as a derivative of a product (e.g., \( \frac{d}{dx}(e^{3x} y) \)).
• Integrate both sides to find a solution in terms of \( y \).

The integration yields a general solution with an arbitrary constant \( C \), which can then be determined using initial conditions. In this exercise, the solution was derived by integrating the transformed equation and using the given conditions to find \( C \), thus resulting in the specific solution.