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 \(t \rightarrow \infty\). $$ t y^{\prime}+2 y=\sin t, \quad t>0 $$

Short answer

In summary, given the first-order linear differential equation \(t y^{\prime}+2 y=\sin t, t>0\), we performed the following steps: 1. Rewrote the equation in the standard form: \(y^{\prime}+ \frac{2}{t} y = \frac{\sin t}{t}, t>0\). 2. Found the integrating factor: \(\mu (t) = e^{2 \ln t} = t^2\). 3. Solved the equation to find the general solution: \(y = \frac{1}{t^2} (-t\cos t + \sin t) + \frac{C}{t^2}\). 4. Analyzed the behavior of the solution as \(t \rightarrow \infty\), observing that the solution tends to 0. 5. Drew a direction field to confirm that the behavior of the solution and the direction field plot agree. The general solution for the given differential equation is \(y = \frac{1}{t^2}(-t\cos t + \sin t) + \frac{C}{t^2}\), and as \(t\) approaches infinity, the solution tends towards 0.

Step-by-step solution

Rewrite the equation in the standard form

We are given the differential equation: $$ t y^{\prime}+2 y=\sin t, \quad t>0 $$ To write it in standard form, divide by \(t\): $$ y^{\prime}+ \frac{2}{t} y = \frac{\sin t}{t}, \quad t>0 $$

Find the integrating factor

Now, to solve the equation, we need to find the integrating factor: $$ \mu (t) = e^{\int \frac{2}{t} \mathrm{d}t} = e^{2 \ln t} = t^2 $$

Solve the equation

Multiply the equation from Step 1 by the integrating factor: $$ t^2 y^{\prime}+2 t y = t^2 \frac{\sin t}{t} $$ Now, the left-hand side is the derivative of \((t^2y)\) with respect to \(t\). So, we can write: $$ \frac{\mathrm{d}(t^2 y)}{\mathrm{d}t} = t \sin t $$ Integrate both sides with respect to \(t\): $$ t^2 y = \int t\sin t \, \mathrm{d}t + C $$ To solve the integral \(\int t\sin t \, \mathrm{d}t\), we'll use integration by parts. Let \(u=t\), \(\mathrm{d}v=\sin t \, \mathrm{d}t\), then \(\mathrm{d}u=\mathrm{d}t\), and \(v=-\cos t\). Applying integration by parts, we get: $$ \int t\sin t \, \mathrm{d}t = -t\cos t + \int \cos t \, \mathrm{d}t\\ = -t\cos t + \sin t + K $$ where \(K\) is the integration constant. Therefore: $$ t^2 y = -t\cos t + \sin t + K $$ Finally, we divide by \(t^2\) to obtain the general solution: $$ y = \frac{1}{t^2} (-t\cos t + \sin t) + \frac{C}{t^2} $$

Analyze the behavior as \(t \rightarrow \infty\)

As \(t \rightarrow \infty\), the terms \(\frac{-t\cos t + \sin t}{t^2}\) and \(\frac{C}{t^2}\) will go to 0, because the exponential growth of the denominator overpowers the terms in the numerator. Therefore, the solution tends towards 0 as \(t \rightarrow \infty\).

Draw a direction field and compare with the solution

We can draw a direction field for the given differential equation using a direction field plotter or software. On the plot, we will see small vectors representing the rate of change of the function at each point. By following the direction field, we can sketch approximations of solution curves. Comparing the direction field plot and the behavior of the general solution, we can confirm that as \(t\) gets larger, the solution curves tend towards 0. This agrees with our analysis in Step 4.

Key concepts

Direction Field

A direction field, also known as a slope field, is a visual representation used to understand differential equations. It provides a graphical way to examine solutions without solving the equation analytically. For a given differential equation, small line segments or arrows are drawn at discrete points in the plane. These vectors characterize the slope of the solution at that specific point.

The direction field provides an intuitive way to see how solutions of the differential equation behave. When you draw the direction field for the equation \(ty^{\prime} + 2y = \sin t\), you use software or manual sketching to mark the slopes. By following the arrows, you can predict the general trends and shapes of the solution curves.

• Shows slope of solutions at each point.
• Helps in predicting behavior without solving the equation.
• Allows visual inspection of solutions' trends.

By examining this direction field, you can start to observe how solutions will behave as \(t\) becomes very large. This insight guides us to more formally solve the equation in subsequent steps.

General Solution

A general solution of a differential equation includes all possible solutions and usually contains arbitrary constants. For our differential equation after rewriting it in standard form \(y^{\prime}+ \frac{2}{t} y = \frac{\sin t}{t},\; t>0\), we applied an integrating factor to solve it.

An integrating factor, \(\mu(t) = t^2\), was found to help convert the differential equation into an easily integrable form. After applying this factor, the equation transforms into the derivative of a product that can be easily integrated:
\[\frac{d}{dt}(t^2 y) = t \sin t\]
The integration of both sides grants us:
\[t^2 y = \int t\sin t \, dt + C\]
This leads us to the general solution:
\[y = \frac{1}{t^2} (-t\cos t + \sin t) + \frac{C}{t^2}\]

Integration by Parts

Integration by parts is a method used to integrate the product of two functions. This method transforms the integral into a simpler form that is easier to evaluate. The formula for integration by parts is given by:
\[\int u \, dv = uv - \int v \, du\]
In our exercise, when solving \(\int t \sin t \, dt\), we decided:

• \(u = t\) and \(dv = \sin t \, dt\)
• The derivatives are \(du = dt\) and \(v = -\cos t\)

Applying integration by parts, we get:
\[\int t\sin t \, dt = -t\cos t + \int \cos t \, dt\]
This solution further led to:
\[-t\cos t + \sin t + K\]
Understanding this method enhances your ability to tackle problems involving product functions in integration. This step is crucial as it allows us to express the integral required for finding the general solution.

Behavior Analysis

Behavior analysis involves studying how the solutions of the differential equation behave as time \(t\) approaches infinity. After finding the general solution
\[y = \frac{1}{t^2} (-t\cos t + \sin t) + \frac{C}{t^2}\]
we can see how each component behaves.

As \(t \to \infty\), both the term \(\frac{-t\cos t + \sin t}{t^2}\) and the constant \(\frac{C}{t^2}\) tend toward zero. This happens because the numerator remains bounded, but the denominator increases without bound, overpowering the numerator.

• The function decays towards zero.
• This aligns with predictions from the direction field.
• Understanding asymptotic behavior is critical for forecasting long-term trends.

Thus, the behavior analysis helps confirm that as \(t\to \infty\), the solutions consistently approach zero. It is consistent with what we observe in the direction field, providing both visual and analytical confirmation.