Question

Determine whether the following equations are separable. If so, solve the initial value problem. $$\sec t y^{\prime}(t)=1, y(0)=1$$

Short answer

Question: Solve the initial value problem \(\sec t y'(t) = 1\) with the initial condition \(y(0) = 1\). Answer: The solution to the initial value problem is \(y(t) = \sin t + 1\).

Step-by-step solution

Check if the equation is separable

We need to rewrite the given equation in the form of \((dy/dt) = h(t)g(y)\). The given equation is: $$\sec t y'(t) = 1$$ Divide both sides by \(\sec t\) to isolate \(y'(t)\): $$y'(t) = \frac{1}{\sec t}$$ Since \(y'(t)\) is already isolated on the left side and there is no \(y(t)\) term on the right side, we can conclude that the equation is separable with \(h(t) = 1/\sec t\) and \(g(y) = 1\).

Integrate both sides to solve the equation

Now that we've determined the equation is separable, we will integrate both sides with respect to \(t\). $$\int y'(t) dt = \int \frac{1}{\sec t} dt$$ The left side's integral is simply \(y(t)\) as the integral of the derivative of \(y\) with respect to time is \(y\) itself: $$y(t) = \int \frac{1}{\sec t} dt$$ To find the integral of the right side, recall that \(\frac{1}{\sec t} = \cos t\). Thus, we have: $$y(t) = \int \cos t dt$$ The integral of \(\cos t\) is \(\sin t\), so we obtain: $$y(t) = \sin t + C$$ where \(C\) is the integration constant.

Utilize the initial condition to find the particular solution

Now, we will use the initial condition \(y(0) = 1\) to determine the value of \(C\). $$y(0) = \sin 0 + C$$ Since \(\sin 0 = 0\), we have \(1 = 0 + C\), which gives us \(C = 1\). Therefore, the particular solution is given by: $$y(t) = \sin t + 1$$ This is the solution to the given initial value problem.

Key concepts

Initial Value Problem in Differential Equations

An initial value problem (IVP) is a type of differential equation that comes with an initial condition. The main task is to find a specific solution that not only satisfies the differential equation but also meets the given initial condition. For example, in the exercise

• The differential equation is \( \sec t y^{\prime}(t) = 1 \).
• The initial condition is \( y(0) = 1 \).

The initial condition provides a starting point or "anchor". It specifies the value of the solution at a particular point, thereby allowing us to determine any constants that arise during integration. In our solution, after solving the differential equation, we found the general solution \( y(t) = \sin t + C \). The initial condition \( y(0) = 1 \) was then used to determine that \( C = 1 \), giving the particular solution \( y(t) = \sin t + 1 \). Initial value problems are essential in modeling real-world scenarios where we know some state or condition at a particular time and want to understand how the system evolves from there.

Integration Techniques for Solving Differential Equations

Integration involves finding a function whose derivative matches a given function. It's a fundamental technique for solving separable differential equations. Once you bring the differential equation into the form of \( \frac{dy}{dt} = h(t)g(y) \), you can integrate both sides appropriately with respect to their variables.In the exercise, the right-hand side of the equation \( y'(t) = \frac{1}{\sec t} \) was integrated. Recognize that \( \frac{1}{\sec t} = \cos t \), simplifying the integral to \( \int \cos t \, dt \).

• Perform the integration: \( \int \cos t \, dt = \sin t + C \)

When integrating, remember to always add a constant of integration, \( C \). This constant represents the family of all possible solutions, from which the initial condition will be used to find the exact solution. Effectively, integration is a means to "reverse" differentiation, extracting the original function when given its rate of change.

Understanding Trigonometric Functions in Differential Equations

Trigonometric functions like \( \sin t \), \( \cos t \), and \( \sec t \) play a crucial role in many differential equations. These functions often appear due to their periodic nature and their relationships with derivatives and integrals.In our specific example, the presence of \( \sec t \) in the original differential equation \( \sec t \, y^{\prime}(t) = 1 \) required manipulation using trigonometric identities.

• \( \sec t = \frac{1}{\cos t} \), leading to \( \frac{1}{\sec t} = \cos t \).

This trigonometric identity transforms the integrand to something more manageable and familiar. Utilizing the integral \( \int \cos t \, dt \) then directly yields \( \sin t \), a primary trigonometric function, helping to solve the equation.Understanding these functions and their properties, such as their integrals and derivatives, is key to solving many problems that involve periodic changes or oscillations, common in physics and engineering applications.