Question

Determine whether the following equations are separable. If so, solve the initial value problem. $$\sec x y^{\prime}(x)=y^{3}, y(0)=3$$

Short answer

Question: Determine the explicit solution of the initial value problem: \(y^{\prime}(x)=\frac{y^{3}}{\sec x}\) with the initial condition \(y(0) = 3\). Answer: The explicit solution to the initial value problem is \(y(x) = \frac{1}{\sqrt{-2(\cos^{-1} x - 19)}}\).

Step-by-step solution

Rewrite the equation

First, we rewrite the given differential equation by isolating the derivative term (\(y^{\prime}(x)\)). $$y^{\prime}(x)=\frac{y^{3}}{\sec x}$$

Check for separable functions

Now, we determine if the equation is separable by seeing if the right-hand side of the equation can be written as a product of functions in x and y. Since \(\sec x\) is a function of x, and \(y^3\) is a function of y, we can rewrite the equation as: $$y^{\prime}(x) = (\cos^{-1} x)(y^{3})$$ This confirms that the equation is separable.

Separate the variables

Next, we separate the variables x and y to different sides of the equation: $$\frac{dy}{y^{3}}=\cos^{-1} x dx$$

Integrate both sides of the equation

Now, we integrate both sides with respect to their respective variables: $$\int\frac{1}{y^{3}}dy=\int\cos^{-1} x dx$$ $$-\frac{1}{2}y^{-2} =\cos^{-1} x + C$$

Solve for y(x)

In this step, we solve for y(x) by performing arithmetic operations: $$y^{-2} = -2(\cos^{-1} x + C)$$ $$y^2 = \frac{1}{-2(\cos^{-1} x + C)}$$ $$y(x) = \pm\frac{1}{\sqrt{-2(\cos^{-1} x + C)}}$$

Apply the initial condition y(0)=3

To find the particular solution, apply the initial condition y(0) = 3: $$3 = \pm\frac{1}{\sqrt{-2(\cos^{-1} 0 + C)}}$$

Solve for the constant C

Now, we determine the value of the constant C: $$3 = \frac{1}{\sqrt{-2(1 + C)}}$$ Squaring both sides, we get: $$9 = \frac{1}{-2(1 + C)}$$ Now, we solve for C: $$-18 = 1 + C$$ $$C = -19$$

Write the final solution

We substitute the constant C back into the solution and use the positive square root (since y is positive when x = 0) to find the final solution to the initial value problem: $$y(x) = \frac{1}{\sqrt{-2(\cos^{-1} x - 19)}}$$

Key concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it involves one or more unknown functions and their derivatives. These equations are crucial in describing various phenomena in fields like physics, engineering, and more.

There are many types of differential equations, but the focus here is on ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives. The order of a differential equation is determined by the highest derivative present. For example, the given equation \( \sec x y'(x) = y^3 \) is a first-order ODE since it involves the first derivative of \( y \).

Understanding differential equations involves recognizing whether they can be easily manipulated or solved using certain techniques, one of which we will explore next.

Initial Value Problem

An initial value problem in the context of differential equations is an equation that requires a solution satisfying a specific condition at a given point. This condition, known as the initial condition, is crucial because it helps determine the unique solution that matches that specific scenario.

In our case, the initial condition is given by \( y(0) = 3 \). This tells us that when \( x=0 \), the function \( y \) takes the value of 3.

With this type of problem, you not only find a general solution for the differential equation, but also adjust your solution to meet this initial condition. By doing so, you narrow down the infinite set of possible solutions to a single one that precisely fits the given criteria.

Integration

Integration is the process of finding functions whose derivative is the given function. As seen in this problem, integration plays a critical role in solving separable differential equations.

The equation is first separated and expressed in terms of differentials: \( \frac{dy}{y^3} = \cos^{-1} x \, dx \).

To solve it, you perform integration on both sides of the equation:

• The left-hand side: \( \int \frac{1}{y^3} \, dy = -\frac{1}{2}y^{-2} \).
• The right-hand side: \( \int \cos^{-1} x \, dx \).

After integrating, an arbitrary constant \( C \) is added to represent general solutions. Much like pieces of a puzzle, integration helps build the function that solves the equation with the given conditions.

Separation of Variables

Separation of variables is a powerful method used to solve ordinary differential equations (ODEs) when they can be written in a form where the variables are on separate sides of the equation.

This technique is particularly useful for first-order separable differential equations such as \( y'(x) = \sec x y^3 \). The goal is to rearrange the equation so that all terms involving one variable (like \( y \)) are on one side, and all terms involving another variable (like \( x \)) are on the opposite side.

Here's how it works for the given equation:

• Recognize \( \frac{d y}{d x} = y^3 \cos^{-1} x\) is separable.
• Rearrange terms to form \( \frac{dy}{y^3} = \cos^{-1} x \, dx \).

This allows us to integrate each side separately, ultimately leading to the solution. By separating the variables, you create a pathway to solving what initially seems an unsolvable problem. This method is straightforward but requires attention to ensure all variables are properly handled.