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Chapter 2: Problem 7

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

Short Answer

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**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

01

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 \(y'\): $$y' - \frac{2+\tan{t}}{\cos{y}} = 0$$
02

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: $$\mu(t) = e^{\int -\frac{\tan{t}}{\cos{y}} dt}$$
03

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: $$-\int \frac{du}{u} = -\ln{|u|} + C = -\ln{|\sec{y}|} + C$$
04

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: $$\mu(t) = \frac{k}{\sec{y}} = k\cos{y}$$
05

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: $$ky' \cos y - k(2+\tan t) = 0$$
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider a population modeled by the initial value problem $$ \frac{d P}{d t}=(1-P) P+M, \quad P(0)=P_{0} $$ where the migration rate \(M\) is constant. [The model (8) is derived from equation (6) by setting the constants \(r\) and \(P_{*}\) to unity. We did this so that we can focus on the effect \(M\) has on the solutions.] For the given values of \(M\) and \(P(0)\), (a) Determine all the equilibrium populations (the nonnegative equilibrium solutions) of differential equation (8). As in Example 1, sketch a diagram showing those regions in the first quadrant of the \(t P\)-plane where the population is increasing \(\left[P^{\prime}(t)>0\right]\) and those regions where the population is decreasing \(\left[P^{\prime}(t)<0\right]\). (b) Describe the qualitative behavior of the solution as time increases. Use the information obtained in (a) as well as the insights provided by the figures in Exercises 11-13 (these figures provide specific but representative examples of the possibilities). $$ M=2, \quad P(0)=0 $$

To establish the uniqueness part of Theorem \(2.1\), assume \(y_{1}(t)\) and \(y_{2}(t)\) are two solutions of the initial value problem \(y^{\prime}+p(t) y=g(t), y\left(t_{0}\right)=y_{0} .\) Define the difference function \(w(t)=y_{1}(t)-y_{2}(t)\). (a) Show that \(w(t)\) is a solution of the homogeneous linear differential equation \(w^{\prime}+p(t) w=0\). (b) Multiply the differential equation \(w^{\prime}+p(t) w=0\) by the integrating factor \(e^{P(t)}\), where \(P(t)\) is defined in equation (11), and deduce that \(e^{P(t)} w(t)=C\), where \(C\) is a constant. (c) Evaluate the constant \(C\) in part (b) and show that \(w(t)=0\) on \((a, b)\). Therefore, \(y_{1}(t)=y_{2}(t)\) on \((a, b)\), establishing that the solution of the initial value problem is unique.

First order linear differential equations possess important superposition properties. Show the following: (a) If \(y_{1}(t)\) and \(y_{2}(t)\) are any two solutions of the homogeneous equation \(y^{\prime}+p(t) y=0\) and if \(c_{1}\) and \(c_{2}\) are any two constants, then the sum \(c_{1} y_{1}(t)+c_{2} y_{2}(t)\) is also a solution of the homogeneous equation. (b) If \(y_{1}(t)\) is a solution of the homogeneous equation \(y^{\prime}+p(t) y=0\) and \(y_{2}(t)\) is a solution of the nonhomogeneous equation \(y^{\prime}+p(t) y=g(t)\) and \(c\) is any constant, then the sum \(c y_{1}(t)+y_{2}(t)\) is also a solution of the nonhomogeneous equation. (c) If \(y_{1}(t)\) and \(y_{2}(t)\) are any two solutions of the nonhomogeneous equation \(y^{\prime}+p(t) y=g(t)\), then the sum \(y_{1}(t)+y_{2}(t)\) is not a solution of the nonhomogeneous equation.

Consider the differential equation \(y^{\prime}=|y|\). (a) Is this differential equation linear or nonlinear? Is the differential equation separable? (b) A student solves the two initial value problems \(y^{\prime}=|y|, y(0)=1\) and \(y^{\prime}=y\), \(y(0)=1\) and then graphs the two solution curves on the interval \(-1 \leq t \leq 1\). Sketch what she observes. (c) She next solves both problems with initial condition \(y(0)=-1\). Sketch what she observes in this case.

Consider a chemical reaction of the form \(A+B \rightarrow C\), in which the rates of change of the two chemical reactants, \(A\) and \(B\), are described by the two differential equations $$ A^{\prime}=-k A B, \quad B^{\prime}=-k A B, $$ where \(k\) is a positive constant. Assume that 5 moles of reactant \(A\) and 2 moles of reactant \(B\) are present at the beginning of the reaction. (a) Show that the difference \(A(t)-B(t)\) remains constant in time. What is the value of this constant? (b) Use the observation made in (a) to derive an initial value problem for reactant \(A\). (c) It was observed, after the reaction had progressed for \(1 \mathrm{sec}\), that 4 moles of reactant \(A\) remained. How much of reactants \(A\) and \(B\) will be left after \(4 \sec\) of reaction time?
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