Problem 27 - Calculus Single Variable

Question

Find the solution of the given initial value problem. $$ y \frac{d y}{d x}=\frac{2 x}{\sqrt{1+y^{2}}} \quad y(0)=0 $$

Step-by-Step Solution

Answer

The implicit solution is \( y^2 = \frac{2x^2}{\sqrt{1+y^2}} \) and satisfies \( y(0) = 0 \).

Step 1: Rewrite the Equation

Rewrite the differential equation given by \( y \frac{dy}{dx} = \frac{2x}{\sqrt{1+y^2}} \) by separating the variables. Multiply both sides by \( dx \), to get: \( y \, dy = \frac{2x}{\sqrt{1+y^2}} \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the separated equation. The left side becomes: \( \int y \, dy = \frac{y^2}{2} + C_1 \). For the right side, multiply by \( \sqrt{1+y^2} \, dx \), yielding: \( \int \frac{2x}{\sqrt{1+y^2}} \, dx \), which is integrated in the next steps.

Step 3: Solve the Right Integral

Set \( z = 1+y^2 \) to simplify \( \int \frac{2x}{\sqrt{1+y^2}} \, dx \). Compute \( dz = 2y \, dy \), so \( \int \frac{2x}{\sqrt{z}} \, dx \) is computed using standard integration, resulting in \( \frac{x^2}{z^{1/2}} + C_2 \).

Step 4: Address the Initial Condition

Given \( y(0) = 0 \), use this to find constants. Substitute \( x=0 \) into equation \( \frac{y^2}{2} = \frac{x^2}{z^{1/2}} + C_2 \) resulting in \( 0 = \frac{0}{\sqrt{1+0}} + C_2 \). This makes \( C_2 = 0 \).

Step 5: Complete the Solution

With \( C_2 =0 \), combine into full equation \( \frac{y^2}{2} = \frac{x^2}{\sqrt{1+y^2}} \) and solve for \( y(x) \). Rewriting gives an implicit function for solution: \( y^2 = \frac{2x^2}{\sqrt{1+y^2}} \).

Step 6: Check Solution and Form

Verify solution matches initial condition \( y(0) = 0 \) and that it is consistent with differential equation. Verify both functional form and condition.

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) consists of a differential equation combined with a set of conditions, called initial conditions. These initial conditions are used to pin down the specific solution out of a family of solutions to a differential equation. Imagine you're given a differential equation—something like a rule that describes a rate of change or relationship between variables. The initial condition provides a starting point, helping you find the one precise solution that fits the scenario.

• A typical IVP includes a differential equation: for example, in our exercise, it's \( y \frac{d y}{d x} = \frac{2 x}{\sqrt{1+y^2}} \).
• An initial condition, like \( y(0) = 0 \), specifies the value of the function at a certain point.

By issuing an initial condition, it becomes possible to integrate the equation and find the specific path that a solution takes through possible values, ensuring a unique function that satisfies both the differential equation and the initial condition.

Separation of Variables

Separation of Variables is a simple integration technique to solve certain differential equations, particularly when the equation can be rearranged to isolate one variable and its derivative on each side. This technique splits the equation, allowing you to solve both parts separately through integration.Begin by looking for parts of the equation that can be moved to each side, separating the variables entirely:

• In our exercise, the expression is first written as \( y \frac{d y}{d x} = \frac{2 x}{\sqrt{1+y^2}} \).
• Multiply both sides by \( dx \) to shift the expressions: \( y \, dy = \frac{2x}{\sqrt{1+y^2}} \, dx \).

With this separation, the path forward involves separate integration of each side, treating each part like a function of a single variable. This clears the way to find the integral, aiming for a function or relationship that satisfies the initial condition.

Integration Techniques

Integration Techniques come into play once you've separated your variables and need to find the antiderivative of each side. This process might require specific approaches based on the forms and complexity of the expressions. Here are some steps and tips:- **Left Side Integration**: Simply, - The left side \( \int y \, dy \) integrates to \( \frac{y^2}{2} + C_1 \). Such a basic integration method is usually applicable when dealing with a simple polynomial form.- **Right Side Integration**: Things can get trickier. - With \( \int \frac{2x}{\sqrt{1+y^2}} \, dx \), consider substitution. Replacing a complex part with a substitute that simplifies the integral, like setting \( z = 1+y^2 \), often helps. - Solve resulting integrals using known integral forms, possibly leading to solutions like \( \frac{x^2}{z^{1/2}} + C_2 \). This process becomes essential when managing functions that aren't straightforward polynomials.Mastering these techniques enables the solving of more challenging integrals, making sense of the differential equation's real-world meanings.

Implicit Function

An Implicit Function is a form where the solution of a differential equation isn't expressed solely as \( y = f(x) \), but instead, variables are combined in a relationship that defines the function relation.In our exercise, after addressing initial conditions and integration, we conclude with:

• \( y^2 = \frac{2x^2}{\sqrt{1+y^2}} \)

This expression is more complicated than a simple \( y = \ldots \) format and does not isolate \( y \) explicitly. Implicit solutions can be directly verified against initial conditions and rewritten to solve particular values, allowing complex relationships without the need to express everything explicitly. Such expressions capture a functional form useful in many scientific and engineering problems where variables inherently depend on each other. Understanding implicit forms broadens your ability to solve and interpret more sophisticated differential equations.