Question

Solve each differential equation. $$ \frac{d y}{d x}-\frac{y}{x}=3 x^{3} ; y=3 \text { when } x=1 $$

Short answer

The solution to the differential equation is \( y = x^4 + 2x \).

Step-by-step solution

Identify the type of differential equation

The given equation is \( \frac{dy}{dx} - \frac{y}{x} = 3x^3 \), which is a first-order linear differential equation. We can express it in the standard form: \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( P(x) = -\frac{1}{x} \) and \( Q(x) = 3x^3 \).

Find the integrating factor

The integrating factor, \( \mu(x) \), is found using \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = -\frac{1}{x} \), so \( \mu(x) = e^{-\int \frac{1}{x} \, dx} = e^{-\ln|x|} = \frac{1}{x} \).

Multiply through by the integrating factor

Multiply the entire differential equation by the integrating factor \( \frac{1}{x} \) to get: \( \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = 3x^2 \).

Simplify to a derivative of a product

Recognize the left side as a derivative of a product: \( \frac{d}{dx}(\frac{y}{x}) = 3x^2 \).

Integrate both sides

Integrate both sides of the equation with respect to \( x \): \( \int \frac{d}{dx}(\frac{y}{x}) \, dx = \int 3x^2 \, dx \). This gives \( \frac{y}{x} = x^3 + C \), where \( C \) is the constant of integration.

Solve for \( y \)

Multiply through by \( x \) to solve for \( y \): \( y = x^4 + Cx \).

Use the initial condition to find \( C \)

Substitute the initial condition \( y = 3 \) when \( x = 1 \) into \( y = x^4 + Cx \) to find \( C \). Plugging the values in gives: \( 3 = 1^4 + C(1) \). Thus, \( C = 2 \).

Write the particular solution

Substitute \( C = 2 \) into the equation \( y = x^4 + Cx \) to get the specific solution: \( y = x^4 + 2x \).

Key concepts

First-Order Linear Differential Equation

In mathematics, understanding differential equations is crucial for modeling many real-world phenomena. A first-order linear differential equation is one of the fundamental types of differential equations you'll encounter. It generally takes the form:

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

Here, the equation features a derivative of the unknown function \( y \) with respect to \( x \), along with terms that possibly involve \( y \) and \( x \) themselves. The coefficients \( P(x) \) and \( Q(x) \) are functions of \( x \).

Key points about first-order linear differential equations:

• They involve the first derivative \( \frac{dy}{dx} \) and no higher derivatives.
• The function \( y \) is linear, meaning it appears without exponents other than 1.
• These equations can often be solved using special techniques, such as integrating factors.

Mastering first-order linear differential equations is an important step in understanding how changes in one quantity can affect another over time or space.

Integrating Factor

An integrating factor is a useful mathematical tool for solving first-order linear differential equations. It simplifies the differential equation, making it easier to solve.

The integrating factor, usually denoted as \( \mu(x) \), is a function you multiply throughout the equation turning the left side into an exact derivative:

\[ \mu(x) = e^{\int P(x) \, dx} \]

In this equation, \( P(x) \) is the coefficient of \( y \) from the standard form of the differential equation. This transformation enables us to rearrange the equation into something that can be integrated more straightforwardly.Steps to use an integrating factor:

• Identify \( P(x) \).
• Calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \).
• Multiply the entire differential equation by \( \mu(x) \).
• The equation now becomes easily integrable on the left side with respect to \( x \).

Utilizing an integrating factor is a powerful technique that turns a seemingly complex equation into a much simpler integral problem.

Initial Condition

In solving differential equations, an initial condition provides the specific values that allow us to find a unique solution. This is necessary because differential equations typically have infinitely many solutions.

An initial condition will usually specify the value of the unknown function \( y \) at a certain point \( x_0 \), such as \( y = 3 \) when \( x = 1 \).How initial conditions work:

• They help in calculating the constant of integration \( C \) after solving the differential equation.
• Using the initial condition, you substitute these values into your general solution.
• Solve the equation for \( C \) to get the particular solution that fits the given condition.

By applying initial conditions, we narrow down the solution to exactly one that suits the physical or theoretical situation described in the problem.