Question

Using the given boundary condition, find the particular solution to each differential equation. Try some by calculator. $$x d x=4 y d y, x=5 \text { when } y=2$$

Short answer

The particular solution to the differential equation is \( \frac{x^2}{2} = 2y^2 + C \) with C calculated from the initial condition to be -6. Therefore, the particular solution is \( \frac{x^2}{2} = 2y^2 - 6 \).

Step-by-step solution

Rewrite the differential equation

Start by rewriting the given differential equation in a separable form. Separate the variables x and y on different sides of the equation.

Integrate both sides

Integrate both sides of the equation with respect to their respective variables. On the left side, integrate with respect to x, and on the right side, integrate with respect to y.

Solve the integrals

Solve the integrals obtained in Step 2. For the left side, the integral of x with respect to x is \( \frac{x^2}{2} \). For the right side, the integral of 4y with respect to y is \( 4\frac{y^2}{2} \) or \( 2y^2 \).

Apply the constant of integration

After integrating, introduce the constant of integration C to one side of the equation. We can choose to add it to the right side for convenience.

Use the initial condition to find C

Apply the initial condition \(x = 5\) when \(y = 2\) to the equation with the constant of integration to solve for the value of C.

Write the particular solution

Substitute the value of C back into the integrated equation to obtain the particular solution to the differential equation.

Key concepts

Separable Differential Equations

When we encounter a differential equation, it often looks like a tangled puzzle waiting to be solved. In our case, the puzzle is a bit easier to handle because we are dealing with a separable differential equation. This type of equation allows us to manipulate the terms so that we can separate the variables on opposite sides of the equation.

Consider the example from the exercise: \( x dx = 4y dy \). To solve this, you act like a magician separating linked rings. By dividing both sides by x and 4y, you can isolate dx on one side and dy on the other, giving you \( \frac{dx}{x} = \frac{dy}{y^2} \). Now, you have set the stage for integrals to enter the scene. You integrate both sides with respect to their own variables. This ease of separation and integration is the charm of separable differential equations! By using the initial condition provided, you finalize the trick and reveal the particular solution.

Integrating Factor

Sometimes equations are more stubborn and won't separate as neatly. When faced with such unruly expressions, we turn to another magic trick - the integrating factor. It's a method used to transform a non-separable equation into a form that can be integrated.

Think of it as a mathematical glue that binds together terms in a way that allows the equation to be integrated more smoothly. For example, if you have a first-order linear differential equation like \( dy/dx + P(x)y = Q(x) \), finding an integrating factor, usually denoted by μ(x) and derived from \( P(x) \), can help you multiply through the equation to make the left side a perfect derivative of y. This integration-friendly format guides you toward the solution.

However, in our initial exercise, the spur for the show was a separable differential equation, where there was no need for an integrating factor. Nonetheless, knowing this technique equips you with a valuable tool for future, more complex battles with differential equations.

Initial Value Problem

When a magician reveals their final trick, the initial setup is what makes the entire act come together. In the realm of differential equations, this setup is known as the initial value problem. It's essentially a puzzle equipped with a starting point. Here, not only do we have a differential equation, but we are also given a specific condition called the initial value that our solution must satisfy.

The initial value serves as an anchor, providing the critical information needed to find the constant of integration after solving the indefinite integrals. For instance, in our exercise, we integrated and found a general solution, but it wasn't until we applied the initial condition \(x = 5 \text{ when } y = 2\) that we could solve for the constant C. This pinpointed the exact function from the family of possible solutions that we were seeking.

By methodically following each step and applying the initial value, we allow our particular solution to elegantly surface, like a card that appears in just the right place at the climax of a magic trick.