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Solve the given differential equation. \(\frac{d y}{d t}=t \cos ^{2} y\)

Short Answer

Expert verified
The solution to the differential equation is \(y = \arctan(0.5t^{2} + C)\) where C is a constant.

Step by step solution

01

Separate the Variables

Rewrite the equation in order to separate the variables, we have: \( dy/\cos^{2}y = td t\)
02

Integrate Both Sides

Now integrate both sides of the equation. The integral of \(1/\cos^{2}y dy\) with respect to y is \(\tan y\). For the right side, the integral of \(tdt\) with respect to t is \(0.5t^{2}\). Hence, we have: \(\tan y = 0.5t^2 + C\) where C is the constant of integration.
03

Solve for y

Taking the arctangent of both sides gives the solution in explicit form: \(y = \arctan(0.5t^2 + C)\)

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Key Concepts

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

Separation of Variables
Differential equations play a vital role in various fields of science and engineering. An important method to solve these equations is the "separation of variables". This technique is used when a differential equation can be rewritten such that each variable and its differential are on opposite sides of the equation.
To solve by separation of variables, follow these simple steps:
  • Identify the differential equation and check if it can be separated.
  • Move all terms involving one variable to one side of the equation, and all terms involving the other variable to the opposite side.
  • Ensure that each side of the equation only contains a single variable and its derivative.
If successful, both the dependent and independent variables should be able to exist on separate sides of the equation, simplifying the process of integration. As seen in the original exercise, the equation \(\frac{dy}{dt} = t \cos^2 y\) was separated to allow easier integration: \(\frac{dy}{\cos^2 y} = t dt\). This separation step allowed for the integration processes to take place conveniently on both sides.
Integration Techniques
Integration is a fundamental operation in calculus and is crucial for solving differential equations. In the method of separation of variables, the next step after separating the variables is to integrate each side of the equation. Different integration techniques might be applied depending on the form of the equation. Let's look at the techniques used in our exercise:

***Integrating Trigonometric Functions***
In this exercise, the integral of \(\frac{1}{\cos^2 y} \, dy\) was evaluated. Recognize that \(\frac{1}{\cos^2 y}\) is the same as \(\sec^2 y\) and integrating \(\sec^2 y\) yields \(\tan y\). Knowing the antiderivatives of trigonometric functions can simplify solving integrals involving them.

***Integrating Polynomial Functions***
On the right-hand side, \( t dt \) represents a simple polynomial function. Integrating \( t \) yields \( \frac{1}{2}t^2 \). Polynomial integrals are typically more straightforward, often involving adding one to the exponent and dividing by the new exponent.
  • Integration transforms a differential equation into a solvable form.
These techniques are some of the most commonly used methods for integration, and mastering them is vital for working with differential equations.
Constant of Integration
The constant of integration is a fundamental concept in calculus. When solving differential equations through integration, an arbitrary constant, typically denoted as \(C\), is added to represent all possible solutions. This is due to the fact that the process of differentiation eliminates constant values.

***Why is the Constant Important?***
When you integrate, there are infinitely many functions that could have been differentiated to yield the same derivative. Thus, the constant \(C\) accounts for all these possibilities. For example, in our solution, after integrating, we found that \(\tan y = 0.5t^2 + C\).

***Determining the Constant***
The value of \(C\) can be determined if an initial condition or additional information is provided. Without such conditions, the solution remains in its general form. To find \(C\),
  • Plug the initial condition into the integrated equation.
  • Solve the equation for \(C\).
Understanding and applying the constant of integration is crucial when solving integrals, especially in the context of initial value problems.

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Most popular questions from this chapter

(a) There are two values of \(\lambda\) such that \(y=e^{\lambda^{t}}\) is a solution to \(y^{\prime \prime}+7 y^{\prime}+12 y=0\). Find them and label them \(\lambda\), and \(\lambda_{2}\). (b) Let \(y=C_{1} e^{\lambda_{1} I}+C_{2} e^{\lambda_{2} t}\), where \(C_{1}\) and \(C_{2}\) are arbitrary constants. Verify that \(y(t)\) is a solution to \(y^{\prime \prime}+7 y^{\prime}+12 y=0\).

Give systems of differential equations modeling competition between two species. In each problem nd the nullclines. The nullclines will divide the phase-plane into regions; nd the direction of the trajectories in each region. Use this information to sketch a phase-plane portrait. Then interpret the implications of your portrait for the long-term outcome of the competition. \(x(t)\) and \(y(t)\) give the number of thousands of animals of species \(A\) and \(B\), respectively. $$ \left\\{\begin{array}{l} \frac{d x}{d t}=0.03 x-0.01 x^{2}-0.01 x y \\ \frac{d y}{d t}=0.05 y-0.01 y^{2}-0.01 x y \end{array}\right. $$

Do a qualitative analysis of the family of solutions to each of the differential equations below. Then, in another color pen or pencil, highlight the graphs of the solutions corresponding to the given initial conditions. (If the solution is asymptotic to a horizontal line, draw and label that line.) \(\begin{array}{llll}\text { (a) } \frac{d y}{d t}=4 y-8 ; & y(0)=0, & y(0)=-1, & y(0)=3\end{array}\) (b) \(\frac{d y}{d t}=y^{2}-4\) \(y(0)=-1, \quad y(0)=-3, \quad y(0)=4\) (c) \(\frac{d y}{d t}=(y-1)(y-2)(y+1) ; \quad y(0)=0, \quad y(0)=3\) \(\begin{array}{llll}\text { (d) } \frac{d y}{d t}=y^{2}+5 y-6 ; & y(0)=-5, & y(0)=-7, & y(0)=2\end{array}\)

Let \(x=x(t)\) be the number of hundreds of animals of species \(A\) at time \(t\). Let \(y=y(t)\) be the number of hundreds of animals of species \(B\) at time \(t\). For each system of differential equations, describe the nature of the interaction between the two species. What happens to each species in the absence of the other? (a) \(\left\\{\begin{array}{l}\frac{d x}{d t}=0.02 x-0.001 x^{2}-0.002 x y \\\ \frac{d y}{d t}=0.008 y-0.004 y^{2}-0.001 x y\end{array}\right.\) (b) \(\left\\{\begin{array}{l}\frac{d x}{d t}=0.02 x-0.01 x y \\ \frac{d y}{d t}=-0.01 y+0.08 x y\end{array}\right.\) (c) \(\left\\{\begin{array}{l}\frac{d x}{d t}=0.02 x-0.001 x^{2}+0.002 x y \\\ \frac{d y}{d t}=0.03 y-0.006 y^{2}+0.001 x y\end{array}\right.\)

Suppose we modify the Volterra predator-prey equations to reflect competition among prey for limited resources and competition among predators for limited resources. The equations would be of the form $$ \left\\{\begin{array}{l} \frac{d x}{d t}=k_{1} x-k_{2} x^{2}-k_{3} x y \\ \frac{d y}{d t}=-k_{4} y-k_{5} y^{2}+k_{6} x y \end{array}\right. $$ where \(k_{1}, k_{2}, \ldots, k_{6}\) are positive constants. Consider the system $$ \left\\{\begin{array}{l} \frac{d x}{d t}=x(1-0.5 x-y) \\ \frac{d y}{d t}=y(-1-0.5 y+x) \end{array}\right. $$ (a) Find the equilibrium points. (b) Do a qualitative phase-plane analysis. (In fact, solution trajectories will spiral in toward the non-trivial equilibrium point.)

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