Question

Solve the following differential equations by using integrating factors. $$ y^{\prime}=2 y-x^{2} $$

Short answer

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

Step-by-step solution

Identify the Form

The given differential equation is \(y' = 2y - x^2\). We rewrite it in the standard linear form: \(y' - 2y = -x^2\).

Determine the Integrating Factor

The standard form for a first-order linear differential equation is \(y' + P(x)y = Q(x)\). Here, \(P(x) = -2\). The integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). So, \(\mu(x) = e^{-2x}\).

Multiply Through by the Integrating Factor

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

Recognize the Left Side as the Derivative of a Product

The left side of the equation can be written as the derivative of a product: \(\frac{d}{dx}(e^{-2x}y) = -x^2e^{-2x}\).

Integrate Both Sides

Integrate both sides with respect to \(x\): \[\int \frac{d}{dx}(e^{-2x}y) \, dx = \int -x^2e^{-2x} \, dx\].This gives us: \[e^{-2x}y = \int -x^2e^{-2x} \, dx + C\].

Solve the Integral on the Right

Use integration by parts for \(\int -x^2e^{-2x} \, dx\). Let \(u = x^2\), \(dv = -e^{-2x} \, dx\). Then, \(du = 2x \, dx\), \(v = \frac{1}{2}e^{-2x}\). Using integration by parts: \[\int u \, dv = uv - \int v \, du\]. This results in \[-\frac{1}{2}x^2e^{-2x} + \int xe^{-2x} \, dx\]. Repeat integration by parts to solve the second integral.

Complete the Integration

For \(\int xe^{-2x} \, dx\), let \(u = x\), \(dv = -e^{-2x} \, dx\). Then, \(du = \ dx\), \(v = \frac{1}{2}e^{-2x}\). This results in \(-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\). The integral becomes:\[-\frac{1}{2}x^2e^{-2x} - \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\].

Solve for \(y\)

Substitute back into the equation for \(e^{-2x}y\): \[e^{-2x}y = -\frac{1}{2}x^2e^{-2x} - \frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + C\].Cancel \(e^{-2x}\) and solve for \(y\):\[y = -\frac{1}{2}x^2 - \frac{1}{2}x - \frac{1}{4} + Ce^{2x}\].

Key concepts

Integrating Factors

Integrating factors are a useful tool for solving first-order linear differential equations. They help transform these equations into an easily integrable form. To understand how integrating factors work, we need to follow a few steps.

The process involves converting the differential equation into its standard linear form, which is generally expressed as \( y' + P(x)y = Q(x) \). From there, we calculate the integrating factor, denoted by \( \mu(x) \), to facilitate the integration process. The integrating factor itself is computed using the exponential function of the integral of \( P(x) \), specifically \( e^{\int P(x) \, dx} \).

By multiplying every term in the original differential equation by this integrating factor, the equation's left side becomes the derivative of a product. This transformation simplifies the integration process, making it more straightforward to solve the differential equation.

First-Order Linear Differential Equations

A first-order linear differential equation is an equation involving derivatives that can be written in a specific form. This form is \( y' + P(x) y = Q(x) \). Here, \( y' \) is the first derivative of \( y \), \( P(x) \) is a function of \( x \), and \( Q(x) \) is another function of \( x \).

These equations are called "first-order" because they involve the first derivative of the function without any higher-order derivatives. Solving these equations typically revolves around isolating \( y' \) and systematically using techniques like integrating factors to simplify the problem.

Understanding how to manipulate these equations into a form that allows us to apply known solving techniques, such as utilizing integrating factors, is crucial. It becomes essential to rewrite the equation in a standard form to make use of the integrating factor method effectively.

Integration by Parts

Integration by parts is a technique often used for solving complex integrals, especially when a product of functions is involved. It is particularly useful in the context of solving integrals that arise in differential equations.

This method is based on the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \]Here:

• \( u \) is a function of \( x \)
• \( dv \) is another function of \( x \) that can be integrated

To apply integration by parts, we carefully choose \( u \) and \( dv \) so that the resulting \( du \) and \( v \) terms simplify the problem.

In the solution outlined, integration by parts was applied more than once. First, \( u \, dv \) was selected based on differentiability, and then the remaining integral was tackled using a repeated application of the same rule. This method leverages manageable derivatives and integrals to simplify the whole integration process.