Question

Solve the following initial-value problems by using integrating factors. $$ y^{\prime}=2 y+x e^{x}, y(0)=-1 $$

Short answer

The solution is \(y = \frac{x^2}{2}e^{2x} - e^{2x}\).

Step-by-step solution

Identify the standard form

The differential equation given is \(y' = 2y + xe^x\). The standard form of a first-order linear differential equation is \(y' + P(x)y = Q(x)\). Here, \(P(x) = -2\) and \(Q(x) = xe^x\).

Calculate the integrating factor

The integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). Here, \(P(x) = -2\), so we compute \(\int -2 \, dx = -2x\). The integrating factor is \(\mu(x) = e^{-2x}\).

Multiply the equation by the integrating factor

Multiply the entire differential equation by the integrating factor \(e^{-2x}\): \(e^{-2x} y' + e^{-2x} (-2y) = xe^x e^{-2x}\). This simplifies to \((e^{-2x} y)' = x\).

Integrate both sides

Integrate both sides of the equation: \(\int (e^{-2x} y)' \, dx = \int x \, dx\), yielding \(e^{-2x} y = \frac{x^2}{2} + C\).

Solve for \(y\)

Isolate \(y\) by multiplying both sides by \(e^{2x}\): \(y = \frac{x^2}{2}e^{2x} + Ce^{2x}\).

Apply the initial condition

Use the initial condition \(y(0) = -1\) to find \(C\): \(-1 = \frac{0^2}{2} + C e^{0}\). This simplifies to \(-1 = C\).

Write the final solution

Substitute \(C = -1\) back into the expression for \(y\). The solution is \(y = \frac{x^2}{2}e^{2x} - e^{2x}\).

Key concepts

First-Order Linear Differential Equations

First-order linear differential equations are a type of differential equation involving the first derivative of a function. They can have a wide range of applications, from predicting population growth to modeling mechanical vibrations.
The general form of a first-order linear differential equation is \( y' + P(x)y = Q(x) \). Here, \( y' \) represents the derivative of \( y \) with respect to \( x \). The functions \( P(x) \) and \( Q(x) \) are known as the coefficients which influence the equation's behavior. In our problem, the equation given was \( y' = 2y + xe^x \). We needed to rearrange it to fit this general form by identifying \( P(x) \) and \( Q(x) \) correctly.
A helpful way to approach these equations is through graphing the approximate solutions. This provides insights into how the solutions behave without solving them analytically. Once organised properly, the task of solving the differential equation becomes more manageable.

Initial-Value Problems

Initial-value problems (IVPs) are a special type of differential equation with specified initial conditions. These initial conditions are crucial because they enable us to find a unique solution that fits specific criteria.
When tackling an IVP, you need to solve the differential equation while ensuring that the solution matches the given initial condition. In our exercise, we were given the initial condition \( y(0) = -1 \). This meant that when \( x = 0 \), the value of \( y \) should be \(-1\).
Understanding the importance of initial values dictates how the solution to the differential equation will be shaped. Initial conditions help to determine the constant of integration, ensuring that the particular solution meets the given criteria exactly. In this particular problem, after applying the initial condition, we found the constant \( C \) to be \(-1\), helping us with a specific solution.

Integrating Factor Method

The integrating factor method is a powerful technique for solving first-order linear differential equations. This method transforms a differential equation into a simpler form, making it easier to solve.
The main idea involves calculating an integrating factor, a function that when multiplied to the original equation, allows it to be rewritten in a form that's straightforward to integrate. For our task, the integrating factor was \( \mu(x) = e^{-2x} \). This was determined by computing \( e^{\int P(x) \, dx} \), where \( P(x) = -2 \).
By multiplying the entire equation by this integrating factor, we simplified the differential equation into \( (e^{-2x} y)' = x \). This form made it possible to integrate both sides easily. After gaining the integrated form, the solution was then adjusted to fit the initial condition, ensuring it was specific to the problem given. The integrating factor method is thus a systematic way to approach and solve differential equations efficiently.