Problem 39 - Calculus and its applications

Question

Solve by separating variables. \(\frac{d y}{d x}=\frac{2 x}{y}\)

Step-by-Step Solution

Answer

After separating and integrating, we have \( y = \pm \sqrt{2x^2 + 2C} \).

Step 1: Rewrite the Differential Equation

Start by rewriting the given differential equation in a form that allows us to separate the variables. The equation is \( \frac{d y}{d x} = \frac{2x}{y} \). Separate the variables by multiplying both sides by \( y \) and by \( dx \): \( y \, dy = 2x \, dx \).

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. The left side with respect to \( y \), and the right side with respect to \( x \): \[ \int y \, dy = \int 2x \, dx \] This yields: \( \frac{y^2}{2} = x^2 + C \), where \( C \) is the integration constant.

Step 3: Solve for y

To express \( y \) in terms of \( x \), multiply through by 2 to eliminate the fraction:\[ y^2 = 2x^2 + 2C \]Then take the square root of both sides, remembering to consider both the positive and negative roots:\[ y = \pm \sqrt{2x^2 + 2C} \].

Key Concepts

Differential Equations

A differential equation is an equation that relates one or more functions and their derivatives. In simpler terms, it connects rates at which things change. For example, instead of telling you directly how something moves, it describes how its speed or direction changes. This type of equation is crucial in science and engineering as it helps in modeling real-world phenomena like population growth, heat conduction, or electrical circuits.
Consider the given differential equation: \[ \frac{d y}{d x} = \frac{2x}{y} \]This equation suggests a relationship between the rate of change of \( y \) in respect to \( x \), and the values of \( x \) and \( y \) themselves. In this case, \( \frac{d y}{d x} \) denotes the derivative of \( y \) with respect to \( x \), showing how \( y \) changes as \( x \) changes.
To work with this, one common technique is separation of variables, ideal for equations where variables can be distinctly divided. This method makes it easier to integrate both sides of the equation separately.

Integration

Integration is the process of finding a function given its derivative. In the world of calculus, this is akin to the opposite of differentiating. While differentiation looks for rates or slopes, integration searches for overall quantities or areas.
In the step-by-step solution, integration is used to find the antiderivative of both sides of the separated equation. After separating variables in the equation:\[ y \, dy = 2x \, dx \]We then integrate both sides:

• For \( y \, dy \), integrating gives \( \int y \, dy = \frac{y^2}{2} \).
• For \( 2x \, dx \), integrating yields \( \int 2x \, dx = x^2 \).

Importantly, after integration, we add a constant \( C \) to the right side to account for the indefinite nature of these integrals. Integrals need this constant since they represent families of functions, and any constant can be added to an antiderivative to modify it slightly but still solve the original differential equation.

Solving Differential Equations

To solve a differential equation, especially those involving separable variables, involves a few systematic steps. Let’s break them down using the equation in the exercise you've encountered.
Start with \( \frac{dy}{dx} = \frac{2x}{y} \). Following the separation of variables, rewrite this as \( y \, dy = 2x \, dx \). This strategy makes the equation easier to handle by allowing us to integrate each side separately:

• On integrating, the left side becomes \( \frac{y^2}{2} \), while the right develops into \( x^2 + C \).

After integrating, further steps involve simplifying to express \( y \) explicitly in terms of \( x \). In our scenario, multiplying through by 2 and considering the square roots gives us the solution: \[ y = \pm \sqrt{2x^2 + 2C} \] This final expression highlights the positive and negative potential outcomes for \( y \), offering a complete solution suite to the differential equation. Solving steps like these require careful arithmetic handling and a good understanding of integration fundamentals.