Question

\((\cos y) y^{\prime}=2+\tan t, \quad y(0)=0\)

Short answer

**Question:** Given the differential equation $(\cos y) y' = 2 + \tan t$ with the initial condition $y(0)=0$, use an integrating factor to solve the equation and identify the function $y(t)$ that satisfies this equation. **Answer:** The given first-order differential equation is solved using an integrating factor, which is determined to be $\mu(t) = \frac{k}{\sec{y}} = k\cos{y}$. The equation is then multiplied by the integrating factor and integrated with respect to $t$. However, the integral may not have a closed-form solution. Some possible techniques to approach the integral include integration by parts or a power series solution.

Step-by-step solution

Rewrite the differential equation in standard form

First, rewrite the given equation as a first-order linear differential equation, dividing through by \(\cos{y}\) to isolate \(y'\): $$y' - \frac{2+\tan{t}}{\cos{y}} = 0$$

Identify the integrating factor

Now that we have the equation in the standard form (\(y' + p(t) y = q(t)\)), identify \(p(t)\) and \(q(t)\): $$p(t) = -\frac{\tan{t}}{\cos{y}}, \quad q(t) = \frac{2}{\cos{y}}$$ To find the integrating factor, compute the integral of \(p(t)\) with respect to \(t\) and exponentiate the result: $$\mu(t) = e^{\int -\frac{\tan{t}}{\cos{y}} dt}$$

Compute the integral

Since we cannot perform the integral directly, we need to apply a substitution. Note that \(\sec y \; dy = dt\). Let \(u = \sec y\), then \(du = \sec y \tan y \; dy\). Making the substitution, we have: $$\int -\frac{\tan{t}}{\cos{y}} dt = \int -\frac{1}{u} \frac{du}{\sec{y}} = -\int \frac{du}{u}$$ Solving this integral, we find: $$-\int \frac{du}{u} = -\ln{|u|} + C = -\ln{|\sec{y}|} + C$$

Exponentiate and find the integrating factor

Now, exponentiate the result to find the integrating factor \(\mu(t)\): $$\mu(t) = e^{-\ln{|\sec{y}|} + C} = e^{-\ln{|\sec{y}|}}e^C = \frac{1}{\sec{y}}e^C$$ For simplicity, let's absorb the constant \(e^C\) into the integrating factor as a new constant, so we have: $$\mu(t) = \frac{k}{\sec{y}} = k\cos{y}$$

Multiply the original differential equation by the integrating factor

Multiply the original differential equation by the integrating factor, \(\mu(t) = k\cos{y}\): $$k\cos y \left(y' - \frac{2+\tan{t}}{\cos{y}}\right) = 0$$ This simplifies to: $$ky' \cos y - k(2+\tan t) = 0$$

Integrate both sides with respect to \(t\)

Now, integrate both sides of the equation with respect to \(t\): $$\int \left(ky' \cos y - k(2+\tan t)\right) dt = \int 0 \; dt$$ In general, this integral may not have a closed-form solution. However, some possible techniques to approach it include using integration by parts or considering a power series solution.

Key concepts

Integrating Factor

An integrating factor is a mathematical tool used for solving first-order linear differential equations. These equations have the form \(dy/dt + p(t)y = q(t)\). A clever trick allows us to transform them into an exact equation that is easy to integrate.

To find the integrating factor, we calculate \(\mu(t) = e^{\int{p(t) \, dt}}\), which is a function of \(t\) only. Multiplying the entire differential equation by this integrating factor makes the left-hand side a derivative of a product. This step consolidates the equation into a simpler form, allowing integration on both sides to find a general solution.

This technique is particularly handy because it requires memorizing formulas and basic steps that work universally for any first-order linear differential equations. The beauty lies in how the integrating factor simplifies the work needed from thereon, making it a must-know tool in differential equation solving.

Substitution Method

The substitution method is a powerful technique used to simplify complex integrals and differential equations. In the context of solving differential equations, like the one given in the exercise, the method involves changing variables to pave an easier path for integration.

For example, if a troublesome term like \(\sec{y}\) appears, you can set \(u = \sec{y}\). This substitution transforms both the equation and the limits of integration into something more manageable. The differential \(du\) is expressed in terms of the new variable, \(u\), and the original variable, \(y\).

• Identify the part of the equation that is complex or repetitive.
• Choose a substitution, like \(u = \sec{y}\).
• Rewrite the differential equation in terms of \(u\).
• Solve the simpler equation, then revert back to the original variable.

By employing substitution, it is possible to bypass complicated calculations, transforming them into straightforward, solvable ones.

Boundary Condition

Boundary conditions are essential elements in differential equations as they provide specific information that allows for the determination of a particular solution to a general differential equation. Essentially, these conditions dictate the value of the solution at a particular point.

In the given exercise, the boundary condition is \(y(0) = 0\). This condition tells you that when \(t = 0\), the value of \(y\) must be zero. Without such a boundary condition, you can only find the general solution that includes arbitrary constants. Applying the boundary condition lets you solve for these constants, thereby finding the exact solution that meets the given criteria.

Boundary conditions are critical in practical applications. They come from physical constraints or special scenarios related to the problem domain, turning general mathematical results into specific real-world solutions.

Trigonometric Identities

Trigonometric identities are the backbone of solving many mathematical problems that involve trigonometric functions. In differential equations, they serve to simplify complex expressions and facilitate both algebraic and calculus-based operations.

Common trigonometric identities include:

• Pythagorean identities like \(\sin^2{\theta} + \cos^2{\theta} = 1\).
• Reciprocal identities such as \(\sec{\theta} = 1/\cos{\theta}\).
• Angle sum identities which allow the breaking down of more complicated functions.

These identities are essential when working with differential equations that feature trigonometric terms. They can simplify integrals or equations, enabling the application of standard calculus techniques.

In this particular exercise, recognizing that \(\tan{t} = \sin{t}/\cos{t}\) could help reframe parts of the equation, thereby revealing simpler paths to integration and further simplifying the solving process.