33–42. Solving initial value problems Solve the following initial value problems.
y'(x) = 4 sec² 2x, y(0) = 8

Verified step by step guidance

1. Identify the given differential equation and initial condition: \(y'(x) = 4 \sec^{2}(2x)\) with \(y(0) = 8\).
2. Recall that to solve the initial value problem, we need to find the antiderivative (integral) of \(y'(x)\) to get \(y(x)\).
3. Set up the integral: \(y(x) = \int 4 \sec^{2}(2x) \, dx + C\), where \(C\) is the constant of integration.
4. Use a substitution to integrate: let \(u = 2x\), so \(du = 2 \, dx\) or \(dx = \frac{du}{2}\). Rewrite the integral in terms of \(u\).
5. Integrate \$4 \sec^{2}(2x) \, dx\( using the substitution, then apply the initial condition \)y(0) = 8\( to solve for \)C$.

Key Concepts

Initial Value Problems (IVPs)

An initial value problem involves a differential equation along with a specified value of the unknown function at a particular point. Solving an IVP means finding a function that satisfies both the differential equation and the initial condition, ensuring a unique solution.

Integration of Trigonometric Functions

Solving the given differential equation requires integrating the derivative function, which involves trigonometric functions like sec²(2x). Recognizing that the integral of sec²(u) du is tan(u) is essential, along with applying substitution when the argument is a function of x.

Applying Initial Conditions to Determine Constants

After integrating, the solution includes an arbitrary constant. Using the initial condition y(0) = 8 allows us to substitute x = 0 and y = 8 into the general solution to solve for this constant, yielding the particular solution to the IVP.