Question

Solve the initial value problem $$ y^{\prime}=2 \cos 2 x /(3+2 y), \quad y(0)=-1 $$ and determine where the solution attains its maximum value.

Short answer

Short Answer: To solve the initial value problem involving the first-order differential equation \(y^{\prime} = \frac{2 \cos 2x}{3 + 2y}\) with the initial condition \(y(0) = -1\), we first separated the variables and integrated both sides. This led to the particular solution \(\frac{3y}{2} + y^2 = \sin 2x + \frac{1}{2}\). To find the maximum value, we set the derivative of y with respect to x equal to zero, which resulted in \(\cos 2x = 0\). The maximum value of the solution is attained at x-values of \(x = \frac{2\pi k + \pi}{4}\), where k is an integer.

Step-by-step solution

Solve the differential equation

First, we need to solve the given differential equation: $$ y^{\prime} = \frac{2 \cos 2x}{3 + 2y} $$ This is a first-order separable differential equation, so we can rewrite it as: $$ (3 + 2y) dy = 2 \cos 2x dx $$ Now, integrate both sides with respect to their respective variables: $$ \int(3 + 2y) dy = \int 2 \cos 2x dx $$

Integrate both sides of the equation

Integrate both sides of the equation to obtain: $$ \frac{3y}{2} + y^2 = \sin 2x + C $$

Apply the initial condition

Now, apply the initial condition \(y(0) = -1\) to find the particular solution: $$ \frac{3(-1)}{2} + (-1)^2 = \sin(2 \cdot 0) + C \\ - \frac{3}{2} + 1 = 0 + C $$ Solving for C, we get: $$ C = \frac{1}{2} $$

Rewrite the particular solution

Now that we have the constant C, we can rewrite the particular solution as: $$ \frac{3y}{2} + y^2 = \sin 2x + \frac{1}{2} $$

Find the maximum value

In order to find the maximum value of the solution, we take the derivative of y with respect to x and set it equal to zero: $$ \frac{dy}{dx} = \frac{2 \cos 2x}{3 + 2y} = 0 $$ Since the only way for the fraction to equal zero is if the numerator is zero, we have: $$ \cos 2x = 0 $$ Thus, the maximum occurs at: $$ x = \frac{2\pi k + \pi}{4} $$ where k is an integer. So, the maximum value of the solution is attained at these x-values.

Key concepts

First-Order Separable Differential Equations

Understanding first-order separable differential equations is essential for solving many problems in calculus. They are equations in which the derivative of the function can be separated into a product of two functions, one that depends only on the independent variable and one that depends only on the dependent variable. In the given exercise, the differential equation can be expressed as:
\( y'=\frac{2\cos(2x)}{3+2y} \).
This equation is separable because it can be manipulated algebraically to isolate the functions of \( y \) and \( x \) on opposite sides of the equation:
\( (3 + 2y) dy = 2\cos(2x) dx \).
By performing this separation, we can then integrate each side individually. Separable differential equations are widely applicable and understanding how to separate and integrate them is a fundamental skill in mathematics.

Integrating Differential Equations

Once we've separated a first-order differential equation, the next step is integrating both sides to find the general solution. During this process, we focus on finding the antiderivative of each side, which often involves standard integration techniques or, at times, substitution and partial fractions. In our example, we are required to integrate:
\( \int (3 + 2y) dy \) and \( \int 2\cos(2x) dx \).
After integration, we add an arbitrary constant \( C \) to one side because indefinite integrals always include an undetermined constant. This gives us the equation:
\( \frac{3y}{2} + y^2 = \sin(2x) + C \).
Remember, integrating is an essential tool not just for differential equations, but for a broad range of applications in fields such as physics, engineering, and economics. Proper integration can help us transition from the rate of change (the derivative) to the quantity itself over a particular interval.

Maximum Value of a Solution

After finding a particular solution to a differential equation, we might be interested in identifying its maximum values. To find the maximum value of a function derived from a differential equation, we look at the derivative of the solution with respect to the independent variable, set it equal to zero, and solve for the variable. This is based on the principle that the slope of a function at a maximum point is zero.
In our exercise, setting the derivative \( \frac{dy}{dx} \) equal to zero:
\( \frac{dy}{dx} = 0 \),
leads to the conclusion that \( \cos(2x) = 0 \) because if the numerator of the derivative is zero, the entire derivative is zero. This allows us to find the values of \( x \) where the maximum occurs, which are given by:
\( x = \frac{2\pi k + \pi}{4} \),
where \( k \) is an integer. By finding these critical points, we can analyze the solution's behavior around these points to confirm they correspond to maximum values. Understanding how to apply calculus to determine the extrema of functions is critical, especially in practical optimization problems, where it is often necessary to maximize or minimize functions related to real-world scenarios.