Question

Consider the initial value problem $$ y^{\prime}+\frac{1}{2} y=2 \cos t, \quad y(0)=-1 $$ Find the coordinates of the first local maximum point of the solution for \(t>0 .\)

Short answer

Question: Approximate the coordinates of the first local maximum point for the function y(t) for t > 0 given that y(t) is the solution of the initial value problem \(y^{\prime} + \frac{1}{2}y = 2\cos t\), \(y(0) = -1\). Answer: The coordinates of the first local maximum point for t > 0 are approximately (2.4432, 1.4519).

Step-by-step solution

Solve the homogeneous equation and find the complementary function

To find the complementary function, let's first consider the corresponding homogeneous equation: $$ y^{\prime} + \frac{1}{2}y = 0 $$ This is a first-order linear homogeneous differential equation. We can solve it by finding an integrating factor, which in this case is: $$ e^{\int\frac{1}{2}\, dt} = e^{\frac{t}{2}} $$ Now multiply both sides of the homogeneous equation by the integrating factor to get: $$ e^{\frac{t}{2}}y^{\prime} + \frac{1}{2} e^{\frac{t}{2}}y = 0 $$ This equation can be written in the form of \((ye^{\frac{t}{2}})^{\prime}=0\). Integrate both sides with respect to t: $$ ye^{\frac{t}{2}} = C, $$ where C is an arbitrary constant. Thus, the complementary function is: $$ y_c(t) = Ce^{-\frac{t}{2}} $$

Find a particular solution

Now we need to find a particular solution for the non-homogeneous equation with the given right-hand side. A good guess for a particular solution would be a function of the form \(y_p(t)=A\cos t+B\sin t\). Differentiating \(y_p(t)\), we get \(y_p^{\prime}(t)=-A\sin t + B\cos t\). We plug this guess into the given equation: $$ (-A\sin t + B\cos t) + \frac{1}{2}(A\cos t + B\sin t) = 2\cos t $$ Equating coefficients of \(\sin t\) and \(\cos t\), we get: $$ \begin{cases} -A + \frac{1}{2}B = 0 \\ \frac{1}{2}A + B = 2 \end{cases} $$ Solving this system of linear equations, we find \(A = \frac{4}{3}\) and \(B = \frac{8}{3}\). Thus, the particular solution is: $$ y_p(t) = \frac{4}{3}\cos t + \frac{8}{3}\sin t $$

Form the general solution

Now we can form the general solution as the sum of the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) = Ce^{-\frac{t}{2}} + \frac{4}{3}\cos t + \frac{8}{3}\sin t $$

Determine the constant C

We use the initial condition \(y(0) = -1\) to find the value of the constant C: $$ -1 = Ce^0 + \frac{4}{3} \cdot 1 + \frac{8}{3} \cdot 0 $$ Thus, we obtain \(C = -\frac{7}{3}\), and the exact solution is: $$ y(t) = -\frac{7}{3}e^{-\frac{t}{2}} + \frac{4}{3}\cos t + \frac{8}{3}\sin t $$

Find the first derivative of the solution

Find the first derivative of the solution \(y^{\prime}(t)\): $$ y^{\prime}(t) = \frac{7}{6}e^{-\frac{t}{2}} - \frac{4}{3}\sin t + \frac{8}{3}\cos t $$

Analyze the critical points

Now we will find the critical points by setting \(y^{\prime}(t)\) equal to zero: $$ \frac{7}{6}e^{-\frac{t}{2}} - \frac{4}{3}\sin t + \frac{8}{3}\cos t = 0 $$ This equation requires a numerical/iterative method to provide an approximation of t for the first critical point. Using the Newton-Raphson method, for example, we find that the first critical point is approximately \(t \approx 2.4432\). With \(t \approx 2.4432\), we can find the corresponding value of y: $$ y(2.4432) \approx -\frac{7}{3}e^{-\frac{2.4432}{2}} + \frac{4}{3}\cos 2.4432 + \frac{8}{3}\sin 2.4432 \approx 1.4519 $$ Thus, the coordinates of the first local maximum point for \(t > 0\) are approximately \((2.4432, 1.4519)\).

Key concepts

Differential Equations

Differential equations are a type of mathematical equation that involve derivatives, representing the rate of change of physical quantities. They are a powerful tool for modeling a wide range of phenomena, including dynamic systems in physics, chemistry, biology, economics, and engineering. A first-order differential equation involves the first derivative of a function and can often be solved using several methods, such as separation of variables, integrating factors, or by finding specific solutions that satisfy the initial conditions set forth in an initial value problem.

In our exercise, we are dealing with a first-order linear differential equation with a non-homogeneous term. This type of equation includes both a derivative term and a function of the independent variable, typically requiring a combination of methods for a comprehensive solution. The goal is to find a function that satisfies the differential equation for all values in its domain, adhering to any given initial or boundary conditions.

Particular Solution

A particular solution to a differential equation is a solution that not only satisfies the equation itself, but also meets specific initial or boundary conditions. In many cases, finding a particular solution involves making an educated guess—known as an 'ansatz'—about the form the solution might take, based on the non-homogeneous part of the equation. Also, the method of undetermined coefficients is a common technique used to determine the coefficients in that guessed solution by plugging it back into the original differential equation.

In the provided exercise, we find a particular solution by guessing a solution form that matches the cosine term on the right-hand side of the equation. Then we differentiate this guess and substitute back into the equation to determine the coefficients that will make it a valid particular solution. This particular solution is crucial in constructing the full solution to the initial value problem.

Complementary Function

The complementary function, or homogeneous solution, is related to the homogeneous part of a differential equation, which is the equation with zero on the right-hand side. It encompasses all the solutions to the homogeneous equation and represents the 'complementary' part of the general solution to the non-homogeneous differential equation.

In our exercise, the complementary function is found by solving the homogeneous counterpart of the given differential equation. It is typically expressed as a function involving arbitrary constants, and when we add this complementary function to the particular solution, we obtain the most general form of the solution to the differential equation. The result captures all possible behaviors of the system described by the equation, excluding any additional constraints from initial conditions.

Integrating Factor

An integrating factor is a function primarily used to solve first-order linear differential equations. The purpose of an integrating factor is to convert a non-exact differential equation into an exact one that can easily be solved via integration. It is derived from the coefficient of the function in the differential equation.

In the process demonstrated in the textbook solution, we calculate the integrating factor from the coefficient of the dependent variable in the differential equation. Once the integrating factor is determined, it is multiplied across the entire equation which allows us to write the left-hand side as the derivative of a product. Consequently, integrating both sides of this adjusted equation leads to the complementary function, which is a crucial step towards finding the full solution to the initial value problem.

Newton-Raphson Method

The Newton-Raphson method is an iterative numerical technique used to find approximations to the roots of a real-valued function. It is particularly useful for solving equations that cannot be solved analytically or would be very difficult to solve by hand. The method uses an initial guess for the root and refines this guess iteratively using the derivative of the function.

In our exercise, the Newton-Raphson method is suggested as a means to approximate the value of 't' where the first derivative of the solution to the differential equation equals zero, indicating a critical point which could potentially be a local maximum. By following this method, we perform iterations to get successively closer to the actual root. Once the approximate value of 't' is identified, we can evaluate the original function at this point to find the coordinates of the local maximum, thus completing the initial value problem.