Problem 26 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ y \frac{d x}{d y}-x=2 y^{2}, \quad y(1)=5 $$

Step-by-Step Solution

Answer

The solution is \( x(y) = 2y^2 - \frac{49}{5}y \) and is defined for all \( y \in (-\infty, \infty) \).

Step 1: Recognize the type of differential equation

The given differential equation is of the form \( y \frac{dx}{dy} - x = 2y^2 \), which can be rewritten as \( y \frac{dx}{dy} = x + 2y^2 \). This suggests that it's a first-order linear differential equation in terms of \( x(y) \).

Step 2: Rearrange the equation

Rearrange the equation to put everything involving \( x \) on one side and the rest on the other: \( \frac{dx}{dy} - \frac{x}{y} = 2y \). This helps set it up for solution methods applicable to linear differential equations.

Step 3: Identify integrating factor

For a first-order linear differential equation \( \frac{dx}{dy} + P(y)x = Q(y) \), the integrating factor \( \mu(y) \) is found using \( \mu(y) = e^{\int P(y) \, dy} \). Here, \( P(y) = -\frac{1}{y} \), so the integrating factor is \( \mu(y) = e^{-\int \frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y} \).

Step 4: Multiply through by the integrating factor

Multiply each term of the equation \( \frac{dx}{dy} - \frac{x}{y} = 2y \) by the integrating factor to get \( \frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = 2 \).

Step 5: Express the left side as a derivative

Rewrite the left-hand side of the equation as a single derivative. This gives us \( \frac{d}{dy} \left( \frac{x}{y} \right) = 2 \). Integrate both sides: \( \int \frac{d}{dy} \left( \frac{x}{y} \right) \, dy = \int 2 \, dy \).

Step 6: Solve the integral

Integrate both sides: \( \frac{x}{y} = 2y + C \). Solve for \( x \): \( x = 2y^2 + Cy \).

Step 7: Apply the initial condition

Use the initial condition \( y(1) = 5 \) to find \( C \). Substitute into the equation: \( 1 = 2(25) + 5C \), which results in \( 1 = 50 + 5C \), giving \( C = -\frac{49}{5} \).

Step 8: Write the solution

Substitute \( C \) back into the equation from Step 6: \( x = 2y^2 - \frac{49}{5}y \).

Step 9: Determine the largest interval \( I \)

The solution \( x(y) = 2y^2 - \frac{49}{5}y \) is a polynomial in \( y \) and thus is defined for all \( y \). Since no restrictions are placed by the equation or initial condition on \( y \), the largest interval \( I \) is \( (-\infty, \infty) \).

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is an equation that involves a function and its derivative, where both are raised to the first power. They have a general form: \( \frac{dy}{dx} + P(x)y = Q(x) \). In this structure, the equation is linear in terms of \( y \) and its derivative.
In our exercise, we began with the equation \( y \frac{dx}{dy} - x = 2y^2 \). By rearranging the terms, we transformed it into \( \frac{dx}{dy} - \frac{x}{y} = 2y \), identifying it as a first-order linear differential equation in terms of \( x(y) \). This format is essential as it leads the way to further solving techniques, like using an integrating factor.
Understanding this type of equation is crucial because it tells us the equation involves both a function and its rate of change but remains linear. This property allows specific methods that simplify finding a solution.

Integrating Factor

An integrating factor is a function used to solve linear first-order differential equations. Its role is to make the equation easily integrable by turning it into a product rule derivative.
For an equation like \( \frac{dx}{dy} + P(y)x = Q(y) \), we calculate the integrating factor \( \mu(y) \) using:

• \( \mu(y) = e^{\int P(y) \, dy} \)

In our case, \( P(y) = -\frac{1}{y} \). The integrating factor becomes \( \mu(y) = \frac{1}{y} \) after evaluating \( e^{\int -\frac{1}{y} \, dy} = \frac{1}{y} \). By multiplying the entire differential equation by this factor, we convert it into \( \frac{d}{dy} \left( \frac{x}{y} \right) = 2 \). This transformed equation can then be straightforwardly integrated.
Using integrating factors effectively transforms differential equations, simplifying the solution process and making complex tasks more manageable.

Solution Interval

The solution interval refers to the range of the independent variable over which the solution to a differential equation is valid. Determining the largest possible interval involves examining the behavior of the solution and any potential restrictions.
Here, our solution \( x(y) = 2y^2 - \frac{49}{5}y \) expressed as a polynomial in terms of \( y \), has no explicit restrictions. Polynomials are defined across all real numbers, indicating the solution exists for any \( y \) value.
The original differential equation and the imposed initial condition \( y(1)=5 \) do not introduce any further constraints on the interval. This analysis reveals that the largest solution interval \( I \) is \( (-\infty, \infty) \). With the absence of singularities or undefined points, we confirm that the entire real line is open for our solution's validity.