Problem 31 - Differential Equations and Linear Algebra

Question

Use the technique derived in the previous problem to solve the given differential equation. $$y^{\prime}+y=e^{-2 x}$$

Step-by-Step Solution

Answer

The general solution to the given first-order linear differential equation \(y' + y = e^{-2x}\) using the Integrating Factor method is \(y(x) = -e^{-2x} + Ce^{-x}\), where C is the integration constant.

Step 1: Determine the Integrating Factor

Recall that for a first-order linear differential equation in the form of \(y' + P(x)y = Q(x)\), the integrating factor (IF) can be obtained as follows: \[IF = e^{\int P(x) dx}\] In our case, the given equaltion is \(y' + y = e^{-2x}\), so we have \(P(x) = 1\). Now let's find the integrating factor: \[IF = e^{\int 1 dx} = e^x\]

Step 2: Multiply both sides of the equation by the Integrating Factor

Now we multiply both sides of the differential equation by \(IF = e^x\): \[e^x(y' + y) = e^x e^{-2x}\]

Step 3: Integrate both sides of the modified equation with respect to x

Notice that the left side of the equation can be simplified as the derivative of the product \(e^x y\): \[\frac{d}{dx}(e^x y) = e^{-x}\] Now, integrate both sides of the equation with respect to x: \[\int \frac{d}{dx}(e^x y) dx = \int e^{-x} dx\] Which simplifies to: \[e^x y = -e^{-x} + C\] Here, C is the integration constant.

Step 4: Solve for y

Finally, we want to solve for y, so we'll divide both sides of the equation by \(e^x\): \[y(x) = -e^{-2x} + Ce^{-x}\] This is the general solution to the given differential equation.

Key Concepts

Integrating Factor

The integrating factor is a powerful tool for solving first-order linear differential equations. For any equation in the form \(y' + P(x)y = Q(x)\), the integrating factor helps transform the equation into a simpler one that can be easily integrated.

To determine the integrating factor, we calculate:

• \(IF = e^{\int P(x)\, dx}\)

In our exercise, the equation \(y' + y = e^{-2x}\) identifies \(P(x) = 1\). This leads us to the integrating factor:

• \(IF = e^{\int 1\, dx} = e^x\)

Once found, the integrating factor is used to multiply through the entire differential equation, simplifying it for further integration. This turns the left side into the derivative of a product, facilitating straightforward integration.

General Solution

The general solution of a differential equation provides a comprehensive expression of the solutions to the equation, capturing all possible situations.

After applying the integrating factor, our modified equation becomes \(\frac{d}{dx}(e^x y) = e^{-x}\). Integration of both sides reveals:

• \(e^x y = \int e^{-x}\, dx = -e^{-x} + C\)

Here, \(C\) represents the integration constant.

Finally, solving for \(y(x)\) gives:

• \(y(x) = -e^{-2x} + Ce^{-x}\)

This expression encompasses all solutions, as different values of \(C\) represent different particular solutions of the differential equation. The flexibility in the choice of \(C\) allows for various initial conditions that might be specified in particular problems.

Differential Equation Solution Techniques

Mastering differential equation solution techniques is essential for handling various types of problems. First-order linear differential equations, like the one given, often require a systematic procedure involving:

• Identifying the standard form \(y' + P(x)y = Q(x)\).
• Determining the integrating factor, \(IF = e^{\int P(x)\, dx}\).
• Applying the integrating factor to simplify the equation.
• Integrating to find the expression for the desired variable.

In our exercise, step-by-step application of these techniques transforms and simplifies the equation effectively. Practicing this approach helps deepen understanding and builds confidence in tackling similar differential equations.