Problem 6 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=\cos ^{2} y \quad\) and \(y=0\) when \(x=0\)

Step-by-Step Solution

Answer

The solution to the initial value problem \(\frac{d y}{d x} = \cos^2(y)\) and \(y = 0\) when \(x = 0\) is \(y = \tan^{-1}(x)\). Its domain is all real numbers except for values of \(x\) at which \(\tan^{-1}(x)\) is undefined. This occurs where \(x = (2n+1)\frac{\pi}{2}\), for \(n\) an integer.

Step 1: Separate the variables

Different parts of the equation involving \(y\) and \(x\) should be isolated on different sides of the equation. Hence, we begin with the equation given: \(\frac{dy}{dx} = \cos^2(y)\). This can be rearranged as \(\frac{dy}{\cos^2(y)} = dx\).

Step 2: Perform the integration

The separated equation can now be integrated term-by-term on both sides. We get: \(\int \frac{dy}{\cos^2(y)} = \int dx\). The integral of \(\frac{1}{\cos^2{y}}\) is \(tan(y)\) and the integral of \(dx\) with respect to \(x\) is \(x\). So the integration solves to \(tan(y) = x + C\).

Step 3: Apply the initial condition

In the equation \(tan(y) = x + C\), let's substitute the initial values \(x = 0\) and \(y = 0\) into the equation. We get: \(tan(0) = 0 + C\). The value of \(tan(0) = 0\), so we can substitute it and solve for \(C\). We have: \(0 = 0 + C\), hence \(C=0\).

Step 4: Substitute the constant C into the solution

Now that we have the value of \(C\) as 0, we can substitute back into the equation from step 2. The solution then is: \(tan(y) = x\).

Key Concepts

Initial Value Problem

An initial value problem in the context of differential equations involves finding a function that not only solves the differential equation but also satisfies an initial condition, which is typically given as the value of the function or its derivatives at a particular point.
For the provided exercise, we have a differential equation \( \frac{dy}{dx} = \cos^2(y) \) combined with an initial condition \( y=0 \) when \( x=0 \) to address. This initial condition is crucial as it will ultimately allow us to determine the specific solution that traces through the point \( (x, y) = (0, 0) \) on the graph of the function, distinguishing it from the general family of solutions to the differential equation.

Integration Techniques

Solving differential equations often requires integrating expressions. The exercise illustrates this step, requiring a technique known as 'separation of variables' to reorganize the equation into a form where each variable and its derivative are on separate sides of the equation.
In our example, the integral \( \int \frac{dy}{\cos^2(y)} = \int dx \) is solved by recognizing \( \frac{1}{\cos^2(y)} \) as the secant squared function, which is the derivative of the tangent function.

Recognizing Standard Integrals

It is often useful to commit to memory the integral of common functions. In our case, knowing that \( \int sec^2(y) dy = \tan(y) + C \) allows for immediate integration without additional steps. This is a testament to the importance of understanding and identifying standard integral forms as a critical integration technique.

Differential Equations

Differential equations are equations that relate a function to its derivatives and are a central part of calculus and other mathematical disciplines. They appear in various forms and complexities, from simple separable equations to more intricate partial differential equations.
The key to solving a differential equation is to understand the relationship it expresses between the rate of change of a quantity and the quantity itself. The exercise provided showcases a relatively straightforward example where the rate of change of \( y \) with respect to \( x \) is given by \( \cos^2(y) \) and shows how separating variables makes it possible to integrate and find the solution.

Applying Initial Conditions

Once the general solution to a differential equation is found using integration techniques, the next step is to apply the initial conditions to determine the specific solution that fits the initial data.
In our equation \( \tan(y) = x + C \) from the exercise, the initial condition \( y(0) = 0 \) allows us to find the particular value of \( C \) that makes the equation match the given point. By substituting \( 0 \) for both \( x \) and \( y \) into the equation, we find that \( C = 0 \).

Interpreting the Specific Solution

This particular solution \( \tan(y) = x \) not only solves the original differential equation but also passes through the point provided by the initial condition, affirming its importance in the overall process of solving an initial value problem.