Problem 23 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ y^{\prime}(x)=x / y(x) \quad y(0)=1 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = \sqrt{x^2 + 1} \).

Step 1: Recognize the Differential Equation Form

The given differential equation is \( y'(x) = \frac{x}{y(x)} \). This is a first-order separable differential equation because it can be rearranged so that one side contains only functions of \( y \) and the other side only functions of \( x \).

Step 2: Separate the Variables

Rearrange the equation to separate variables: \( y \, dy = x \, dx \). This looks like terms involving \( y \) on one side and terms involving \( x \) on the other side.

Step 3: Integrate Both Sides

Integrate both sides of the equation: \( \int y \, dy = \int x \, dx \). The left side integrates to \( \frac{y^2}{2} + C_1 \) and the right side to \( \frac{x^2}{2} + C_2 \). However, combining the constants gives us: \( \frac{y^2}{2} = \frac{x^2}{2} + C \).

Step 4: Solve for \( y \)

Multiply the whole equation by 2 to eliminate fractions: \( y^2 = x^2 + C \). Solve for \( y \) to get \( y = \pm \sqrt{x^2 + C} \).

Step 5: Apply Initial Condition

Use the initial condition \( y(0) = 1 \) to find \( C \). Substitute \( x = 0 \) and \( y = 1 \) into \( y = \pm \sqrt{x^2 + C} \) to get \( 1 = \pm \sqrt{0^2 + C} \), which gives \( C = 1 \) and chooses the positive root \( y = \sqrt{x^2 + 1} \) because \( y(0) = 1 > 0 \).

Step 6: Write the Final Solution

The solution to the initial value problem is \( y(x) = \sqrt{x^2 + 1} \).

Key Concepts

First-Order Differential Equation

A first-order differential equation involves derivatives of an unknown function with respect to a single variable. In such equations, the highest order of the derivative that appears is one. For our exercise, the differential equation is presented as \( y'(x) = \frac{x}{y(x)} \). Here, \( y'(x) \) represents the first derivative of \( y \) with respect to \( x \). First-order differential equations can come in different forms, such as linear or separable, each requiring specific techniques to solve them.

When dealing with these equations, identifying the type is crucial because it dictates the method of solution. In this exercise, the equation is separable, which strongly influences our approach, as we'll see in later sections.

Initial Value Problem

An initial value problem in differential equations requires not only finding a general solution but also determining a particular solution that satisfies an initial condition. This means we are given a point on the solution curve, typically formatted as \( y(0) = 1 \), meaning \( y(x) \) has the value 1 when \( x \) is 0. This condition helps us find the exact solution that fits the specific criteria given.

• Providing an initial value helps to locate the specific solution among the family of possible solutions.
• The initial condition is applied after you find the general solution of the differential equation.
• Solving the initial value problem ensures the uniqueness of the solution curve that matches the given conditions.

It acts as an additional piece of information and is crucial for understanding how the function behaves.

Integration

Integration is a mathematical process used to solve differential equations, particularly after separating variables. It's essentially the reverse process of differentiation. After separating the variables in the equation \( y \, dy = x \, dx \), you integrate both sides with respect to their respective variables. This gives:

• \( \int y \, dy = \frac{y^2}{2} + C_1 \)
• \( \int x \, dx = \frac{x^2}{2} + C_2 \)

Combining the constant terms can simplify the results to \( \frac{y^2}{2} = \frac{x^2}{2} + C \).

• Integration helps in finding the function when its derivative is known.
• It converts the differential equation into an algebraic form which is simpler to handle.
• It is essential for determining the relationship between the variables once they have been separated.

Subsequently, integration paves the way towards solving for the specific function \( y \) and applying initial conditions.

Separation of Variables

The process of separation of variables is a method used to simplify and solve first-order differential equations like the one we are dealing with in this exercise. The method involves rearranging the equation so that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side.

In our example, starting from \( y'(x) = \frac{x}{y(x)} \), you multiply both sides by \( y(x) \) and \( dx \) to get \( y \, dy = x \, dx \). This separation allows each side to be integrated independently:

• Separation helps stratify the problem into simpler, solvable parts; it converts a complex problem into one that can be tackled using basic calculus.
• By integrating each side, we find general solutions for each variable that we can manipulate to find specific solutions.
• This technique is powerful as it is applicable to a wide range of problems in mathematical analysis.

The core of separation of variables lies in its ability to reduce the complexity of differential equations by treating each part independently.