Question

Solve the following initial-value problems by using integrating factors. $$ x y^{\prime}=y-3 x^{3}, y(1)=0 $$

Short answer

The solution is \( y = \frac{3}{2}x - \frac{3}{2}x^3 \).

Step-by-step solution

Rewrite as a Standard Linear Differential Equation

First, rewrite the given differential equation in the standard linear form. The given equation is: \[ x y^{\prime} = y - 3x^3 \]Divide the entire equation by \(x\) to isolate \(y^{\prime}\):\[ y^{\prime} - \frac{y}{x} = -3x^2 \] This is now in the standard form \( y^{\prime} + P(x)y = Q(x) \), where \(P(x) = -1/x\) and \(Q(x) = -3x^2\).

Find the Integrating Factor

Find the integrating factor \( \mu(x) \) using the formula: \[ \mu(x) = e^{\int P(x) \, dx} \] Substitute \(P(x) = -1/x\):\[ \mu(x) = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = |x|^{-1} = \frac{1}{x} \]

Multiply Through by the Integrating Factor

Multiply each term of the differential equation by the integrating factor \( \frac{1}{x} \):\[ \frac{1}{x}(y^{\prime} - \frac{y}{x} ) = \frac{1}{x}(-3x^2) \]Simplify:\[ \frac{dy}{dx} - \frac{y}{x^2} = -3x \]

Solve the Modified Equation

Recognize that the left-hand side is now a derivative of a product:\[ \frac{d}{dx}(\frac{y}{x}) \]Thus:\[ \frac{d}{dx}(\frac{y}{x}) = -3x \]Integrate both sides with respect to \(x\):\[ \int \frac{d}{dx}(\frac{y}{x}) \, dx = \int -3x \, dx \]\[ \frac{y}{x} = -\frac{3}{2}x^2 + C \]

Solve for y

Multiply through by \(x\) to solve for \(y\):\[ y = -\frac{3}{2}x^3 + Cx \]

Apply the Initial Condition

Use the initial condition \( y(1) = 0 \) to solve for \(C\):\[ 0 = -\frac{3}{2}(1)^3 + C(1) \]\[ C = \frac{3}{2} \]Substitute \(C\) back into the equation for \(y\):\[ y = -\frac{3}{2}x^3 + \frac{3}{2}x \]

Final Solution

The solution to the initial-value problem is:\[ y = \frac{3}{2}x - \frac{3}{2}x^3 \]

Key concepts

Integrating Factors

An integrating factor is a mathematical tool used to solve linear differential equations. Its main goal is to make the differential equation easier to solve. For a differential equation in the form of \[ y^{\prime} + P(x)y = Q(x) \]we look for a function, known as the integrating factor \( \mu(x) \), which when multiplied by every term in the equation, transforms the left side into the derivative of a product. This new equation thus becomes easier to integrate. The integrating factor is defined as:\[ \mu(x) = e^{\int P(x) \, dx} \]This is calculated by taking the exponential of the integral of \(P(x)\). By meticulously applying the integrating factor, we convert the differential equation into a format that can be solved more straightforwardly through direct integration.

Linear Differential Equations

Linear differential equations are a fundamental type of differential equation essential in various scientific fields, from physics to engineering. These equations have two main features: the unknown function and its derivatives are linearly associated. In general terms, a first-order linear differential equation can be represented as:\[ y^{\prime} + P(x)y = Q(x) \]Here, \(P(x)\) and \(Q(x)\) are functions of \(x\), and the term \(y^{\prime}\) represents the derivative of \(y\) with respect to \(x\). The linear relationship ensures that powers of \(y\) and \(y^{\prime}\) are no more than one, simplifying the solution process. Understanding these equations is crucial because they frequently arise in modeling real-world phenomena, such as population growth, heat transfer, and electrical circuits.

Integrating Factor Method

The integrating factor method is a crucial technique to solve linear differential equations of the form:\[ y^{\prime} + P(x)y = Q(x) \]The method works in these key steps:

• Identify the functions \(P(x)\) and \(Q(x)\) in your differential equation.
• Find the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the whole equation by the integrating factor \(\mu(x)\).
• Observe that the left side of the equation represents the derivative of a product \( \frac{d}{dx}(\mu(x)y) \).
• Integrate both sides to find \(\mu(x) y(x)\) and solve for \(y(x)\).

This technique effectively transforms a potentially complex problem into a manageable one by exploiting the properties of derivatives.

Initial Conditions

Initial conditions are essential for finding a unique solution to a differential equation. They provide specific information about the function at a given point. In initial-value problems, these conditions typically appear as\[ y(x_0) = y_0 \]where \(x_0\) is a specific value of \(x\), and \(y_0\) is the value of the function at \(x_0\). Applying the initial condition ensures that the solution to the differential equation is not just a general solution but one that satisfies the given initial point, tailoring the solution to a particular scenario. In problems involving standard linear differential equations, the initial condition plays a pivotal role after finding the general solution, helping to determine the specific constant involved, thus providing the precise solution applicable to the problem context.