Problem 23 - Calculus Single Variable

Question

In each of Exercises \(17-28,\) solve the given initial value problem. $$ \frac{d y}{d x}-y=e^{x} \quad y(0)=-3 $$

Step-by-Step Solution

Answer

The solution is \( y = e^{x}(x - 3) \).

Step 1: Identify the Type of Differential Equation

The equation \( \frac{d y}{d x} - y = e^{x} \) is a first-order linear differential equation in standard form \( \frac{d y}{d x} + P(x)y = Q(x) \), where \( P(x) = -1 \) and \( Q(x) = e^{x} \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) is given by \( e^{\int P(x) \, dx} = e^{-\int 1 \, dx} = e^{-x} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor: \( e^{-x} \frac{d y}{d x} - e^{-x}y = e^{x}e^{-x} \). This simplifies to \( e^{-x} \frac{d y}{d x} - e^{-x}y = 1 \).

Step 4: Recognize the Left Side as a Derivative

The left side of the equation \( e^{-x} \frac{d y}{d x} - e^{-x}y \) can be written as \( \frac{d}{dx} (e^{-x} y) \). Thus the equation becomes \( \frac{d}{dx} (e^{-x} y) = 1 \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx} (e^{-x} y) \, dx = \int 1 \, dx \Rightarrow e^{-x} y = x + C \].

Step 6: Solve for \( y \)

Solve for \( y \) by multiplying both sides by \( e^{x} \): \[ y = e^{x}(x + C) \].

Step 7: Apply the Initial Condition

Use the initial condition \( y(0) = -3 \) to find \( C \):\[ -3 = e^{0}(0 + C) \Rightarrow -3 = C \].

Step 8: Write the Particular Solution

Substitute \( C = -3 \) back into the equation for \( y \):\[ y = e^{x}(x - 3) \].

Key Concepts

First-order Linear Differential Equation

A first-order linear differential equation is a relationship that involves an unknown function, say \(y(x)\), its first derivative \( \frac{d y}{d x} \), and a given function of \(x\). Such equations are crucial because they describe a range of phenomena in fields like physics and engineering. The standard form is given by:

• \( \frac{d y}{d x} + P(x)y = Q(x) \)

The equation \( \frac{d y}{d x} - y = e^{x} \) fits into this format. Here, we have \(P(x) = -1\) and \(Q(x) = e^{x}\).
First-order means the highest derivative is of order one, which makes it convenient for applications requiring simple calculations and interpretations.

Integrating Factor

The integrating factor is a crucial tool for solving first-order linear differential equations. It transforms a non-exact differential equation into something that can be written as a derivative of a product of functions. For the equation \( \frac{d y}{d x} - y = e^{x} \), the integrating factor \(\mu(x)\) is determined by:

• \( \mu(x) = e^{\int P(x) \, dx} = e^{-x} \)

This function simplifies the differential equation significantly:

• When \( \frac{d y}{d x} - y = e^{x} \) is multiplied by \(e^{-x}\), the left side turns into a recognizable derivative which is easier to integrate.

Without this simplifying step, solving would be much more complex.

Particular Solution

The particular solution of a differential equation is the specific solution that satisfies an initial condition. When a solution is obtained, we often incorporate conditions or constraints that further define it. In this exercise:

• After finding the general solution \( y = e^{x}(x + C) \), we use the initial value provided, \(y(0) = -3\), to solve for \(C\).
• Substituting \(C = -3\) leads to the particular solution: \(y = e^{x}(x - 3)\).

This solution fully describes the behavior of the system modeled by the differential equation given the initial value.

Initial Condition

Initial conditions are crucial in determining a particular solution for a differential equation. They are values given for the function and/or its derivatives at a specific point, and they distinguish one solution from many possible others.

• In our problem, the initial condition is \( y(0) = -3 \), which means when \(x\) is 0, \(y\) should be -3.

This initial information is invaluable as it allows us to find \(C\) in our general solution \( y = e^{x}(x + C) \).
Consequently, with \(C\) found, we determine the specific path the function takes, ensuring the solution is tailored to meet the condition of the problem precisely.