91Ó°ÊÓ

A separable differential equation is one where the variables can be separated on opposite sides of the equation. This means we can isolate all the terms containing the dependent variable (let's say, "y") on one side and all the terms involving the independent variable (say, "t") on the other side. This separation allows us to integrate each side independently to find the solution.

For example, consider the equation from our original problem, given as:

• \( y'(t) = \cos^2 y \)

Here, the equation is already in a separable form because "t" is not directly present on the right side. By rewriting it, we can separate the variables as:

• \( \frac{dy}{dt} = \cos^2 y \)
• \( \frac{dy}{\cos^2 y} = dt \)

A primary challenge in solving separable differential equations is identifying when this separation of variables is possible. Once separated, integrating each side becomes the task to find the solution.

Trigonometric substitution is a technique often used to simplify integrals involving trigonometric functions. This method involves substituting a trigonometric identity into the integral to make it easier to solve.

In the given problem, once we have the equation \( \int \frac{dy}{\cos^2 y} = \int dt \), we use trigonometric substitution to simplify the left-hand side. Recognizing that \( \sec^2 y = \frac{1}{\cos^2 y} \), we set:

• \( u = \tan y \)
• \( du = \sec^2 y \ dy = \frac{dy}{\cos^2 y} \)

This substitution turns the left side into a simpler integral:

• \( \int du = u \)

Substitutions like this transform complex expressions into simpler algebraic ones, making them easier to integrate.

Integrating differential equations is often the final step in solving them. Once we have our variables separated and substitution applied, we move on to integrating both sides.

For our equation, after substitution:

• Left side: \( \int du = u \)
• Right side: \( \int dt = t + C_1 \)

Here, "C_1" is a constant of integration, introduced because indefinite integrals can vary by a constant.

After integration, the solution is expressed in terms of the original variables. By substituting back \( u = \tan y \), we arrive at \( \tan y = t + C_1 \). Remember to apply the initial condition \( y(1) = \frac{\pi}{4} \) to find the specific value of the constant "C_1" which exhibits the particular solution within the context of the problem.

Integrating differential equations using these steps helps us arrive at the complete solution, reaffirming our understanding of both calculus and differential equations.