Question

(a) Draw a direction field for the given differential equation. (b) Based on an inspection of the direction field, describe how solutions behave for large \(t .\) (c) Find the general solution of the given differential equation and use it to determine how solutions behave as \(t \rightarrow \infty\). $$ y^{\prime}+(1 / t) y=3 \cos 2 t, \quad t>0 $$

Short answer

Answer: The general solution to the given differential equation is \(y(t) = \frac{3}{5}\cos 2t + \frac{3}{10t}\sin 2t + \frac{C}{t}\), where \(C\) is an arbitrary constant. Question: How do the solutions behave as \(t \rightarrow \infty\)? Answer: As \(t \rightarrow \infty\), the solutions approach a fixed value with oscillations that have decreasing amplitude, eventually stabilizing to the limiting value.

Step-by-step solution

Draw the direction field

To draw the direction field, we need to create a grid of points in the \((t, y)\) plane and draw small line segments at each point with their slopes defined by the given equation $$ y' + \frac{1}{t}y = 3 \cos 2t . $$ Plot these small line segments on the grid to see how the solutions behave. Loading the differential equation into a direction field plotter such as https://www.geogebra.org/m/Ap99b54Z will help you create the graph.

Analyze the direction field for large \(t\) behavior

Upon inspection of the direction field, we can see that the solutions appear to oscillate while tending to a limiting value as \(t\) becomes large. The amplitude of the oscillations decreases as \(t\) increases, implying that solutions stabilize to a fixed value as \(t \rightarrow \infty\).

Solve the given differential equation

To solve the differential equation, we can use an integrating factor. The integrating factor, \(u(t)\), satisfies the equation $$ u'(t) = u(t) \cdot \frac{1}{t}. $$ Integrating both sides we get $$ \ln(u(t)) = \ln(t^1) + C_1, $$ so $$ u(t) = kt. $$ for some constant \(k\) (without loss of generality, we can set \(k=1\), so \(u(t) = t\)). Now multiply the given equation by the integrating factor \(u(t) = t\), we have $$ ty^\prime + y = 3t\cos 2t. $$ Thus, the left-hand side is a perfect derivative: $$ \dfrac{d}{dt}(ty) = 3t\cos 2t. $$ Integrate both sides with respect to \(t\): $$ ty = \int 3t\cos 2t dt + C. $$ To obtain the general solution, we can look up this integral in a table or use the integration-by-parts formula. The result is: $$ ty = \frac{3}{5}t\cos 2t + \frac{3}{10}\sin 2t + C. $$ So, the general solution is given by, $$ y(t) = \frac{3}{5}\cos 2t + \frac{3}{10t}\sin 2t + \frac{C}{t}. $$

Analyze the general solution for \(t \rightarrow \infty\)

From the general solution, we can see that as \(t\) becomes very large, the second term \(\frac{3}{10t}\sin 2t\) and the third term \(\frac{C}{t}\) both go to zero, leaving only the first term, \(\frac{3}{5}\cos 2t\). However, the amplitude of the oscillations is still decreasing as \(t\) gets large. Thus, the limiting behavior of solutions as \(t \rightarrow \infty\) is an oscillation with decreasing amplitude, which eventually approaches a fixed value. This observation aligns with the description of the solutions' behavior from the direction field visualization.

Key concepts

Direction Field

A direction field, also known as a slope field, is a visual representation of a differential equation. It involves plotting a grid of points and at each point drawing a small line segment with a slope given by the differential equation evaluated at that point. For our differential equation,

\(y' + \frac{1}{t}y = 3 \cos 2t\), we map out these segments on the \((t, y)\) plane. By examining the direction field, we gain insight into the behavior of solutions without actually solving the equation. In this case, it suggests oscillatory behavior, and as \(t\) gets larger, the solutions appear to stabilize, indicating that the long-term solution may approach a steady state.

Integrating Factor

The integrating factor is a mathematical tool used to solve certain types of first-order linear differential equations. It transforms a non-exact differential equation into an exact one, making it possible to find the general solution. The integrating factor,
\(u(t)\), depends on the differential equation and, for our equation, is found to be \(u(t) = t\). When we multiply the original equation by \(u(t)\), the left-hand side becomes the derivative of \(ty\), leading to an equation that is easier to integrate. This method is highly systematic, making it a powerful technique when encountering first-order linear differential equations. It's also a crucial part of understanding how to approach and solve such problems.

General Solution

The general solution to a differential equation encapsulates all possible specific solutions. It includes an arbitrary constant (or constants) because it represents a family of solutions, rather than a single solution. For our exercise, the general solution has the form
\(y(t) = \frac{3}{5}\cos 2t + \frac{3}{10t}\sin 2t + \frac{C}{t}\), where \(C\) is an arbitrary constant. This expression is the result of integrating the modified equation after applying an integrating factor. It's vital to recognize the significance of the general solution; it indicates the behavior of the system under various initial conditions. Exploring how different values of \(C\) affect the solution further enhances our understanding of the system described by the differential equation.

Oscillatory Behavior

Oscillatory behavior in a differential equation's solution refers to any solution that exhibits repeated fluctuations above and below a central value, much like the sine and cosine functions. The given differential equation contains a cosine term, which hints at the oscillatory nature of its solutions. As \(t\) increases, the terms involving \(t\) in the denominator approach zero, causing the oscillation amplitude to decrease. In our problem, the general solution shows that while the oscillation continues indefinitely, the presence of the \(\frac{1}{t}\) term ensures that the amplitude of the oscillation diminishes as \(t\) becomes large, signifying a damping effect. Eventually, the movement tends toward a steady state, albeit still oscillating with an ever-decreasing magnitude.