To solve the initial value problem:

$t^{-1} \frac{dy}{dt} = 4 \cos^2 y, \quad y(1) = \frac{7\pi}{4}$

we start by separating the variables. First, rewrite the differential equation:

$\frac{dy}{dt} = 4 t \cos^2 y$

Separate the variables $t$ and $y$:

$\frac{1}{\cos^2 y} dy = 4t dt$

Note that $\frac{1}{\cos^2 y} = \sec^2 y$:

$\sec^2 y \, dy = 4t \, dt$

Integrate both sides:

$\int \sec^2 y \, dy = \int 4t \, dt$

The integral of $\sec^2 y$ is $\tan y$, and the integral of $4t$ is $2t^2$:

$\tan y = 2t^2 + C$

Now, apply the initial condition $y(1) = \frac{7\pi}{4}$:

$\tan\left(\frac{7\pi}{4}\right) = 2(1)^2 + C$

Since $\frac{7\pi}{4} = \frac{2\pi}{4} + \pi = \frac{5\pi}{4}$, and $\tan\left(\frac{7\pi}{4}\right) = \tan\left(\frac{-\pi}{4}\right) = -1$:

$-1 = 2 + C$

Solve for $C$:

$C = -3$

So the solution to the differential equation is:

$\tan y = 2t^2 - 3$

The implicit solution is:

$\tan y - 2t^2 = -3$

Or equivalently:

$\tan y = 2t^2 - 3$

Would you like more details or have any questions?

Here are 5 questions you might ask:

1. Can you explain the integration process in more detail?
2. What are some common initial value problems and their solutions?
3. How do we handle trigonometric functions in differential equations?
4. Can you show another example of solving a differential equation with initial conditions?
5. What are some techniques for solving non-separable differential equations?

Tip: Always apply initial conditions after integrating to determine the constant of integration accurately.