Question

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

Short answer

The solution is \( y = 4e^x - 2(x^2 + 2x + 2) \).

Step-by-step solution

Identify the Standard Form

The given differential equation is \( y' = y + 2x^2 \). This can be rewritten in the standard linear differential equation form: \( y' - y = 2x^2 \), where \( P(x) = -1 \).

Calculate the Integrating Factor

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

Multiply the Differential Equation by the Integrating Factor

Multiply each term of the equation \( y' - y = 2x^2 \) by the integrating factor \( e^{-x} \):\[ e^{-x} y' - e^{-x} y = 2x^2 e^{-x} \]

Recognize the Left Side as a Derivative

The left side of the equation \( e^{-x} y' - e^{-x} y \) simplifies to the derivative of \( e^{-x}y \):\[ \frac{d}{dx}(e^{-x} y) = 2x^2 e^{-x} \]

Integrate Both Sides

Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(e^{-x} y) \, dx = \int 2x^2 e^{-x} \, dx \]The left side integrates to \( e^{-x} y \), while the right side requires integration by parts.

Perform Integration by Parts on the Right Side

For \( \int 2x^2 e^{-x} \, dx \), use integration by parts twice:- Let \( u = x^2 \) and \( dv = 2e^{-x} \, dx \).- Then, \( du = 2x \, dx \) and \( v = -2e^{-x} \).- Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \), then repeat for the remaining integral.This eventually gives \( -2e^{-x}(x^2 + 2x + 2) + C \), where \( C \) is an integration constant.

Solve for \( y \)

We have:\[ e^{-x} y = -2e^{-x}(x^2 + 2x + 2) + C \]Multiply through by \( e^x \) to solve for \( y \):\[ y = -2(x^2 + 2x + 2) + Ce^x \]

Apply the Initial Condition

Using the initial condition \( y(0) = 0 \), substitute \( x = 0 \) and \( y = 0 \) into the equation:\[ 0 = -2(0^2 + 2\cdot0 + 2) + C \cdot e^0\]\[ 0 = -4 + C \]Thus, \( C = 4 \).

Write the Final Solution

Substitute \( C = 4 \) back into the expression for \( y \):\[ y = -2(x^2 + 2x + 2) + 4e^x \]This simplifies to:\[ y = 4e^x - 2(x^2 + 2x + 2) \]

Key concepts

Initial-Value Problems

Initial-value problems are a common type of problem in differential equations. They involve finding a function, often denoted as \( y(x) \), that not only satisfies a given differential equation but also meets specific initial conditions. These initial conditions can include values of the function or its derivatives at certain points. This allows us to find a unique solution tailored to the context described by the initial conditions. In the problem provided, we have the initial condition \( y(0) = 0 \), meaning the solution must pass through the point \((0,0)\) on the \( xy\)-plane. A pivotal aspect of solving initial-value problems is accurately applying the initial conditions to determine any constants that arise during the integration process. This ensures the solution is specific and meaningful.

Linear Differential Equations

Linear differential equations are a class of equations characterized by the presence of a dependent variable and its derivatives, which appear linearly. This means the dependent variable and its derivatives do not have powers greater than one and are not multiplied by each other. The standard form for a first-order linear differential equation is \( y' + P(x)y = Q(x) \). Here, \( P(x) \) and \( Q(x) \) are functions of the independent variable \( x \).
The equation given in the exercise is initially in the form \( y' = y + 2x^2 \). By manipulation, it is rewritten as \( y' - y = 2x^2 \), aligning it with the linear form where \( P(x) = -1 \) and \( Q(x) = 2x^2 \). Solving such equations often requires techniques like finding an integrating factor, which helps simplify the process and find solutions efficiently.

Integration by Parts

Integration by parts is a powerful technique used to solve integrals that are products of functions, especially when the standard methods of integration don't apply easily. The method is grounded in the integration by parts formula:

• \( \int u \, dv = uv - \int v \, du \)

It essentially turns a complex integral into simpler parts that are more manageable to solve.
In the given problem, we encounter the integral \( \int 2x^2 e^{-x} \, dx \) on the right-hand side. This requires applying integration by parts multiple times to handle the quadratic polynomial and the exponential function effectively. By selecting \( u \) and \( dv \) correctly, the complexity of the integral can be reduced, allowing for a step-by-step unraveling that eventually provides the necessary integrated expression needed to move forward with solving the equation.

Integrating Factor Method

The integrating factor method is an elegant approach for solving linear first-order differential equations. This technique involves multiplying the entire differential equation by a specially chosen function, known as the integrating factor.

• The form: \( \mu(x) = e^{\int P(x) \, dx} \).

This factor is designed to convert the left-hand side of the equation into the derivative of a product, specifically, the derivative of \( \mu(x)y \).
In our exercise, \( P(x) = -1 \), leading to the integrating factor \( \mu(x) = e^{-x} \). Multiplying through by this factor simplifies the equation to a form where we recognize the left side as \( \frac{d}{dx}(e^{-x} y) \). This maneuver simplifies the integration process and is key to finding the solution to initial-value problems. Once integrated, solving for \( y\) and applying the initial condition completes the problem-solving process, providing an exact and tailored solution.