17–32. Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem.
y'(t) = cos² y, y(1) = π/4

Verified step by step guidance

1. First, rewrite the differential equation in Leibniz notation: \(\frac{dy}{dt} = \cos^{2} y\).
2. Check if the equation is separable by expressing it as a product of a function of \(y\) and a function of \(t\). Here, rewrite as \(\frac{dy}{dt} = \cos^{2} y = f(y) \cdot g(t)\), where \(f(y) = \cos^{2} y\) and \(g(t) = 1\).
3. Since the equation is separable, separate variables by dividing both sides by \(\cos^{2} y\) and multiplying both sides by \(dt\): \(\frac{1}{\cos^{2} y} dy = dt\).
4. Integrate both sides: \(\int \frac{1}{\cos^{2} y} dy = \int dt\). Recall that \(\frac{1}{\cos^{2} y} = \sec^{2} y\), and the integral of \(\sec^{2} y\) with respect to \(y\) is \(\tan y\).
5. After integrating, apply the initial condition \(y(1) = \frac{\pi}{4}\) to solve for the constant of integration and express the solution implicitly or explicitly in terms of \(y\) and \(t\).

Key Concepts

Separable Differential Equations

A differential equation is separable if it can be written as a product of a function of t and a function of y, allowing the variables to be separated on opposite sides of the equation. This form enables integration with respect to each variable independently to find the solution.

Initial Value Problems (IVP)

An initial value problem involves solving a differential equation with a given initial condition, such as y(t₀) = y₀. This condition helps determine the specific solution curve among the family of solutions by fixing the constant of integration.

Integration Techniques for Trigonometric Functions

Solving differential equations involving trigonometric functions often requires using identities and integration methods, such as rewriting cos² y using power-reduction formulas. Mastery of these techniques is essential to integrate and solve the equation explicitly.