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Chapter 2: Problem 4

Find the solution to the indicated initial value problem, and use ezplot to plot it. \(y^{\prime}=-y^{2} \cos t\) with \(y(0)=3\) over \([0,3]\).

Short Answer

Expert verified
The solution is \( y = \frac{3}{1 + 3 \sin t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = -y^2 \cos t \). This is a first-order, separable differential equation because it can be expressed in the form \( g(y)y' = h(t) \).
02

Separate Variables

Separate the variables by dividing both sides by \( y^2 \) and multiplying both sides by \( dt \), leading to \( \frac{dy}{y^2} = -\cos t \, dt \).
03

Integrate Both Sides

Integrate both sides with respect to their respective variables. The left integrates to \( \int \frac{dy}{y^2} = - \frac{1}{y} + C_1 \) and the right integrates to \( \int -(\cos t) \, dt = - \sin t + C_2 \). Thus, the equation becomes \( - \frac{1}{y} = - \sin t + C \), where \( C = C_2 - C_1 \).
04

Solve for the Constant of Integration

Using the initial condition \( y(0) = 3 \), substitute \( t = 0 \) and \( y = 3 \) into the equation \( - \frac{1}{y} = - \sin t + C \) to find \( C \). This gives \( - \frac{1}{3} = - \sin(0) + C \), so \( C = -\frac{1}{3} \).
05

Solve for y(t)

Now substitute \( C = -\frac{1}{3} \) back into the equation: \( -\frac{1}{y} = - \sin t - \frac{1}{3} \). Rearrange to get \( y = \frac{3}{1 + 3 \sin t} \).
06

Plot the Solution

To plot the solution \( y = \frac{3}{1 + 3 \sin t} \) over the interval \([0, 3]\), use a graphing tool such as MATLAB's ezplot. The command may look like `ezplot('3/(1 + 3*sin(t))', [0, 3])`. This will provide a visual representation of the solution over the specified interval.

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a fundamental type of differential equations where variables can be separated on opposite sides of the equation. This allows us to integrate each side individually and find a solution. In a scenario with a given equation like \( y' = -y^2 \cos t \), we rearrange it as \( \frac{dy}{y^2} = -\cos t \, dt \).
This shows how the equation is separated into two integrable parts:
  • \( \frac{dy}{y^2} \) on the left side represents the dependent variable \( y \).
  • \(-\cos t \, dt \) on the right side represents the independent variable \( t \).
The ability to independently integrate these parts streamlines the process of finding a solution. This method is especially useful when initial or boundary conditions are involved, as it simplifies finding specific solutions tailored to the conditions given.
Initial Value Problems
An initial value problem involves finding a specific solution to a differential equation that satisfies a given condition at a particular point. For example, in our exercise, the initial condition is \( y(0) = 3 \). This means that when \( t = 0 \), the value of \( y \) must be 3.
By using this initial condition, we can determine the constant of integration that arises during the integration process. In our example:
  • After integration, the general solution becomes \( -\frac{1}{y} = -\sin t + C \).
  • Applying the initial condition \( y(0) = 3 \) helps us calculate \( C \), leading to \( C = -\frac{1}{3} \).
This step refines the general solution into a specific one, \( y = \frac{3}{1 + 3 \sin t} \), which is applicable over the interval defined in the problem.
Graphing Differential Equations
Graphing the solution of a differential equation is an excellent way to visualize its behavior over a given interval. For the exercise addressed, our task was to graph \( y = \frac{3}{1 + 3 \sin t} \) over the interval \([0, 3]\). By plotting this equation, several observations can be made:
  • We can easily see how \( y \) changes as \( t \) increases from 0 to 3.
  • The plot provides insight into the nature of the solution, revealing trends or specific behavior like oscillations linked to the \( \sin t \) term in the denominator.
Graphing is crucial for verifying analytical results and understanding qualitative traits of the solution, such as stability or periodic tendencies that might not be immediately obvious from the equation itself.
MATLAB
MATLAB is a powerful tool for solving and visualizing complex mathematical problems, including ordinary differential equations. Specifically, for our solution, MATLAB's `ezplot` function is used to graph the solution \( y = \frac{3}{1 + 3 \sin t} \) over the interval \([0, 3]\).
Here's how MATLAB is used in this context:
  • The command `ezplot('3/(1 + 3*sin(t))', [0, 3])` is executed to generate a plot of the function.
  • This provides a clear and precise visual representation, allowing us to see the behavior of \( y \) relative to \( t \) within the specified range.
MATLAB helps in testing and exploring different scenarios and observing impacts immediately, making it an indispensable tool for engineers and scientists working with differential equations.

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Most popular questions from this chapter

Find the general solution of the differential equation. Then plot the family of solutions with the indicated initial values over the specified interval. We will use MATLAB notation to indicate the range of initial values. You can use the method of Example 7, but think about using a for loop. \(y^{\prime}=y-3 e^{-t}\) on the interval \([-2,2]\) with initial conditions \(y(0)=-5: 1: 5\)

Use the appropriate sum-to-product identity from trigonometry to show that $$ \sin (12 t)-\sin (14 t)=-2 \sin (t) \cos (13 t) $$ On one plot, plot \(y=-2 \sin (t) \cos (13 t)\) and its envelopes \(y=\pm 2 \sin (t)\) over the time interval \([-2 \pi, 2 \pi]\). Use the selection tool on the figure toolbar to select each envelope, then right-click the selected envelope to change both its color and linewidth.

If the Symbolic Toolbox is installed in your MATLAB system, use the dsolve command to find the solution of the first order initial value problems in Exercises 15 - 18. Use the ezplot command to plot the solution over the indicated interval. y^{\prime}=-2 t y, y(0)=1,[-2,2]

The voltage across the capacitor in a driven \(R C\)-circuit is modeled by the initial value problem \(V_{c}^{\prime}+V_{c}=\) \(\cos (t), V_{c}(0)=0\). The solution of the problem can be written \(V_{c}=V_{\mathrm{t}}+V_{\mathrm{s}}\), where $$ V_{\mathrm{s}}=\frac{1}{2} \cos (t)+\frac{1}{2} \sin (t) \quad \text { and } \quad V_{\mathrm{t}}=-\frac{1}{2} e^{-t} $$ The solution \(V_{\mathrm{s}}\) is called the steady-state solution and the solution \(V_{\mathrm{t}}\) is called the transient solution. On one plot, sketch the solutions \(V_{\mathrm{s}}, V_{\mathrm{t}}\), and \(V_{c}\) in blue, red, and black, respectively, over the time interval \([0,6 \pi]\). Add a legend to your plot.

For each set of parametric equations in Exercises \(29-32\), use a script file to create two plots. First, draw a plot of both \(x\) and \(y\) versus \(t\). Use handle graphics to apply different linestyles and color to the plots of \(x\) and \(y\), then add a legend, axis labels, and a title. Open a second figure window by placing the \(f\) igure command at this point in your script, then draw a plot of \(y\) versus \(x\). Add axis labels and a title to your second plot. \(x=\cos (2 t)-8 \sin (2 t), y=-5 \sin (2 t)-\cos (2 t),[0,4 \pi]\)
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