91Ó°ÊÓ

Separable differential equations are a type of first-order differential equation where the variables can be separated onto different sides of the equation. This means we can rewrite the equation such that all terms involving one variable are on one side, while terms involving the other variable are on the opposite side.
For the given problems, this allows us to integrate each side separately, making it simpler to solve. Here is a general approach:

• Start with a differential equation in the form \( \frac{dy}{dt} = g(t)h(y) \).
• Rearrange to get \( h(y) dy = g(t) dt \).
• Integrate both sides to solve for \( y \) in terms of \( t \).

In Problem (a), the equation \( \frac{dy}{dt} = \frac{3}{\sqrt{1-t^2}} \) was already mostly separated, making the integration direct after multiplying both sides by \( dt \). In Problem (b), separating got us to work with \( \int \left(1 - \frac{2}{x^2+1}\right) \, dx \), which required a partial fractions approach to tackle effectively. Understanding how to separate and adjust the differential equation lays the groundwork for solving these types of problems.

Integration is a fundamental technique used to solve differential equations once we have isolated the variables. For our problems, specific integration techniques were employed to find solutions.
In Problem (a), recognizing the integrand was crucial. The integral \( \int \frac{1}{\sqrt{1-t^2}} \, dt \) is a trigonometric integral, specifically associated with the inverse sine function. The antiderivative of \( \frac{1}{\sqrt{1 - t^2}} \) gives \( \sin^{-1}(t) \), which we scaled by 3 due to multiplication in the differential equation.
In Problem (b), partial fraction decomposition came into play. This technique involves expressing a complex rational function as a simpler sum of fractions, making them easier to integrate. Here, recognizing \( \int \frac{1}{x^2+1} \, dx \) leads to \( \tan^{-1}(x) \), another inverse trigonometric function integral. By breaking down \( \frac{x^2 - 1}{x^2 + 1} \) into \( 1 - \frac{2}{x^2+1} \), we simplified the integral to solve it directly.

Inverse trigonometric functions are crucial when solving many integrals in differential equations. They help to revert relationships that are not directly algebraic.
These functions have specific derivatives, which allow us to recognize their integrals quickly. Some common inverses include:

• \( \sin^{-1}(x) \)
• \( \cos^{-1}(x) \)
• \( \tan^{-1}(x) \)

In Problem (a), knowing that \( \sin^{-1}(t) \) is the antiderivative of \( \frac{1}{\sqrt{1-t^2}} \) was key to finding \( y(t) \).
For Problem (b), \( \tan^{-1}(x) \) is the antiderivative of \( \frac{1}{x^2+1} \). Recognizing this helped solve the integral \( \int \frac{1}{x^2+1} \, dx \), leading to \( y(x) \). These inverse functions simplify integrals by connecting them to familiar geometric angles.