Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d y}{d x}=\frac{x}{y} ; y=1\) at \(x=1\)

Step-by-Step Solution

Answer

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

Step 1: Separate Variables

Start by separating variables in the differential equation \( \frac{d y}{d x} = \frac{x}{y} \). This means rearranging the equation to get one variable on each side, resulting in \( y \, dy = x \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation to find the general solution. Integrate \( \int y \, dy \) and \( \int x \, dx \). This gives \( \frac{y^2}{2} = \frac{x^2}{2} + C \), where \( C \) is the integration constant.

Step 3: Solve for the General Solution

Multiply everything by 2 to simplify the equation: \( y^2 = x^2 + 2C \). This represents the general solution of the differential equation.

Step 4: Apply Initial Condition

Use the initial condition \( y = 1 \) at \( x = 1 \) to find \( C \). Substitute \( y = 1 \) and \( x = 1 \) into \( 1^2 = 1^2 + 2C \). This simplifies to \( 0 = 2C \), meaning \( C = 0 \).

Step 5: Write the Particular Solution

Substitute \( C = 0 \) back into the general solution \( y^2 = x^2 + 2C \) to get \( y^2 = x^2 \). Solving this gives the particular solution \( y = x \) or \( y = -x \). Considering the positive initial condition \( y = 1 \) when \( x = 1 \), the appropriate solution is \( y = x \).

Key Concepts

General Solution

The general solution of a differential equation is a representation of all possible solutions that a differential equation can have. It embodies a family of curves or functions with one or more arbitrary constants, typically denoted by a symbol like \( C \). For the given differential equation \( \frac{d y}{d x} = \frac{x}{y} \), the process to find the general solution involves separating the variables, as seen in the first step of the solution.
After separation, we combine all terms involving \( y \) on one side and \( x \) on the other, yielding \( y \, dy = x \, dx \).
By integrating both sides, we are able to obtain the expression \( \frac{y^2}{2} = \frac{x^2}{2} + C \). This is our general solution before we apply any specific conditions.

• Integration includes finding antiderivatives, which introduces the integration constant \( C \).
• Each value of \( C \) represents a different curve from the family of solutions.

By multiplying both sides by 2, we further simplify the equation to \( y^2 = x^2 + 2C \). This equation now represents the general solution for the differential equation.

Particular Solution

A particular solution of a differential equation results by substituting a specific condition into the general solution. It represents a single, unique solution curve, among the family of solutions described by the general solution.
In this exercise, the particular solution is based on the initial condition provided, which is \( y = 1 \) when \( x = 1 \).

• Substitute these values into the general solution \( y^2 = x^2 + 2C \) to find the specific \( C \).
• Doing this results in \( 0 = 2C \), leading to \( C = 0 \).
• Substituting \( C = 0 \) back into the general solution gives \( y^2 = x^2 \), which simplifies to two potential solutions, \( y = x \) or \( y = -x \).

Given the initial condition \( y = 1 \) when \( x = 1 \), the suitable particular solution is \( y = x \). It satisfies the initial condition and represents a specific path from our family of solutions.

Separation of Variables

Separation of variables is a fundamental technique for solving differential equations. It involves rearranging the equation so that all terms involving one variable (say \( y \)) are on one side while terms involving the other variable (\( x \)) are on the opposite side. This sets the stage for straightforward integration.
In our exercise, the original differential equation was \( \frac{d y}{d x} = \frac{x}{y} \). The separation technique led to \( y \, dy = x \, dx \).

• Each side of the equation only contains one variable, allowing independent integration.
• This method simplifies the integration process, making it easier to solve the equation analytically.

After separation, integrating each side gives us integral forms that lead smoothly toward the general solution. This step is critical for transforming complex differential equations into more manageable expressions.

Initial Condition

An initial condition in a differential equation problem is a specified value for the function and often its derivatives at a particular point. It is essential to find a particular solution from the general solution.
In this exercise, the initial condition given is \( y = 1 \) when \( x = 1 \). When we plug these values into the general solution equation \( y^2 = x^2 + 2C \), it immediately helps determine the constant \( C \).

• Substituting the initial values gives \( 1^2 = 1^2 + 2C \), simplifying to \( 0 = 2C \), and hence \( C = 0 \).
• An initial condition ensures that the solution corresponds to the conditions in the actual problem context.
• Solving with this specific condition refines the general solution to describe a unique, specific curve, which in this case leads us to \( y = x \).

These conditions give a real-world context or a specific setting to the mathematical solution, honing down from infinite possibilities to a singular viable path.