Question

Show that \(y=\sin t\) is a solution of $$ y^{\prime \prime}+\left(k \sin ^{2} t\right) y^{\prime}+(1-k \cos t \sin t) y=0 $$ for any value of the constant \(k .\) If \(00\) and \(k \sin ^{2} t \geq 0\). Thus observe that even though the coefficients of this variable coefficient differential equation are nonnegative (and the coefficient of \(y^{\prime}\) is zero only at the points \(t=0, \pi, 2 \pi, \ldots\), it has a solution that does not approach zero as \(t \rightarrow \infty .\) Compare this situation with the result of Problem \(38 .\) Thus we observe a not unusual situation in the theory of differential equations: equations that are apparently very similar can have quite different properties.

Short answer

Question: Prove that \(y = \sin t\) is a solution of \((y'' + k \sin^2 t y' + (1 - k \cos t \sin t)y = 0\) for any constant \(k\). Additionally, show that for \(0 < k < 2\), two conditions hold true which are \(1 - k \cos t \sin t > 0\) and \(k \sin^2 t \geq 0\), and compare the behaviour of the solution as \(t \rightarrow \infty\) with Problem 38. Answer: By finding the first and second derivatives of \(y = \sin t\) and plugging them into the given differential equation, we proved that \(y = \sin t\) is a solution for any value of constant \(k\). We also showed that the conditions \(1 - k \cos t \sin t > 0\) and \(k \sin^2 t \geq 0\) hold true for \(0 < k < 2\). Comparing the behavior of this solution as \(t \rightarrow \infty\) with Problem 38, we observed that \(y = \sin t\) oscillates between -1 and 1, while the solution in Problem 38 approaches 0 under specific conditions. This illustrates that closely related differential equations can exhibit different properties and behaviors.

Step-by-step solution

First and Second Derivative

First, differentiate \(y(t) = \sin t\) with respect to \(t\): $$ y^{\prime}(t) = \frac{d}{dt}(\sin t) = \cos t $$ Next, differentiate \(y'(t) = \cos t\) with respect to \(t\): $$ y^{\prime \prime}(t) = \frac{d}{dt}(\cos t) = -\sin t $$

Substitute Derivatives

Plug the derivatives of \(y\) into the given differential equation: $$ (-\sin t) + \left(k \sin^2 t\right) (\cos t) + \left(1 - k \cos t \sin t\right) (\sin t) = 0 $$

Rewrite and Simplify

Rewrite and simplify the equation: $$ -\sin t + k \cos t \sin^3 t + \sin t - k \cos t \sin^2 t = 0 $$ Combining like terms: $$ k \cos t \sin^3 t - k \cos t \sin^2 t = 0 $$ Factor out \(k \cos t \sin^2 t\): $$ k \cos t \sin^2 t(\sin t - 1) = 0 $$

Prove Conditions

As \(k \cos t \sin^2 t(\sin t - 1) = 0\), \(y = \sin t\) is a solution of the given differential equation for any value of constant \(k\). Now we will prove the two conditions: 1. \(1-k\cos t \sin t>0\): This condition is true when \((\sin t - 1) > 0\). We know that the sine function oscillates between -1 and 1; however, it never takes the value of 1. Therefore, this condition holds true for all values of \(t\). 2. \(k\sin^2t\geq0\): Since \(0 < k < 2\), it is a positive value, and \(\sin^2t\) will always be non-negative (ranging from 0 to 1). Thus, the product of a positive \(k\) and \(\sin^2t\) will be greater than or equal to \(0\).

Solution's Behaviour as \(t \rightarrow \infty\) and Comparison

As \(t\) goes to infinity, \(y = \sin t\) oscillates between -1 and 1, and it does not approach 0. In the case of Problem 38, the solution approaches 0 as \(t \rightarrow \infty\) when the coefficients are positive (and under other conditions). This shows how closely related differential equations can have different behaviours and properties, as seen in this exercise.

Key concepts

Understanding Variable Coefficient Differential Equations

Variable coefficient differential equations are fascinating and complex mathematical expressions used to describe a variety of dynamic systems. Unlike their constant coefficient counterparts, these equations have coefficients that change with respect to the independent variable, often making them more challenging to solve. The exercise provided delves into a specific kind of variable coefficient differential equation, with coefficients dependent on trigonometric functions.

In this context, we see the equation \(y^{\text{\prime\prime}} + (k \sin^{2} t)y^{\text{\prime}} + (1 - k\cos t\sin t)y = 0\), where \(k\) is a constant, and the coefficients include sine and cosine terms with variable \(t\). The primary characteristic that defines this as a variable coefficient differential equation is the presence of the \(\sin^{2} t\) and \(\cos t\sin t\) terms that vary with time \(t\). This variation leads to unique solutions and behaviors, as opposed to what we'd expect from equations with constant coefficients.

The exercise proves that \(y = \sin t\) is a solution for any value of \(k\), which isn't necessarily intuitive. Normally, we might expect such a solution to approach zero as \(t\rightarrow\infty\), indicative of a stabilizing system. Yet, our sine function solution oscillates indefinitely, showcasing how equations of this nature can defy our initial expectations and emphasizing the need for robust analytical or numerical methods to understand them fully.

Delving into Second-order Differential Equations

Second-order differential equations, as seen in our textbook exercise, are crucial in the realm of physics and engineering, representing many natural phenomena such as motion, electricity, and heat transfer. These equations involve the second derivative of the unknown function, which suggests a dependency on acceleration or curvature in its physical interpretation.

In the exercise, the equation \(y^{\text{\prime\prime}} + (k \sin^{2} t)y^{\text{\prime}} + (1 - k\cos t\sin t)y = 0\) is a second-order differential equation. The presence of the second derivative, denoted by \(y^{\text{\prime\prime}}\), indicates that the solution \(y\) is a function whose rate of change (velocity) and the rate of change of the rate of change (acceleration) are both relevant to the system's behavior. These equations often require two initial conditions to find a unique solution, corresponding to the initial state and initial rate of change of the system.

Second-order differential equations can be categorized based on the nature of their coefficients. They can be linear or nonlinear, homogenous or nonhomogeneous, and as in this case, they may involve constant or variable coefficients. Each type requires specific methods for finding solutions, and the insight gained from solving them is profound as they directly relate to how systems evolve over time.

Solving Differential Equations: A Fundamental Skill

Solving differential equations is a cornerstone of applied mathematics, underpinning many scientific and engineering disciplines. The process involves finding a function or a set of functions that satisfy the given differential equation. This exercise illustrates the intricate process of solving a variable coefficient differential equation, revealing how even a straightforward, systematic approach can unearth complex behaviors.

The process follows several steps. First, the derivatives of the proposed solution, \(y = \sin t\), are found. Then, these derivatives are substituted back into the original differential equation. The equation is then simplified, leading to an equation that must hold true under certain conditions, validating the proposed solution.

In this case, we evaluated the properties of the coefficients and the behavior of the solution, particularly for values of \(k\) within specific bounds. Such analysis is crucial, as it informs us about the nature of the solutions we can expect – oscillatory, as with \(\sin t\), or convergent to some limit. Solution behavior can vary widely, even among seemingly similar differential equations, which is aptly demonstrated by comparing the given problem to Problem 38 highlighted in the exercise.

It's important for students to grasp the techniques for solving various types of differential equations, as well as understanding the implications of any solution found. This understanding allows for the application of differential equations to real-world scenarios, predicting and analyzing complex system dynamics, and engineering solutions that are reliable and efficient.