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Consider the solution of the differential equation \(y^{\prime}=-3 y\) passing through \(y(0)=1.5\). A. Sketch the slope field for this differential equation, and sketch the solution passing through the point (0,1.5) B. Use Euler's method with step size \(\Delta x=0.2\) to estimate the solution at \(x=0.2,0.4, \ldots, 1\), using these to fill in the following table. (Be sure not to round your answers at each step!) \begin{tabular}{|l|l|l|l|l|l|l|} \hline\(x=\) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline\(y \approx\) & 1.5 & & & & & \\ \hline \end{tabular} C. Plot your estimated solution on your slope field. Compare the solution and the slope field. Is the estimated solution an over or under estimate for the actual solution? D. Check that \(y=1.5 e^{-3 x}\) is a solution to \(y^{\prime}=-3 y\) with \(y(0)=1.5\).

Step by step solution

01

Sketch the Slope Field

To sketch the slope field for the differential equation \(y' = -3y\), evaluate the slope \(y'\) at several points in the (x, y) plane. For example, when \(y = 1\), the slope \(y' = -3\), and when \(y = -1\), the slope \(y' = 3\). Draw short line segments at these points representing the slopes. Continue this process for different values of y.

02

Sketch the Solution Curve

After creating the slope field, sketch the solution curve that passes through the initial condition (0, 1.5). This curve should follow the direction of the slope segments in the field.

03

Euler's Method - Initial Point

Euler's method starts at the initial point (0, 1.5). The formula used is \(y_{n+1} = y_n + f(x_n, y_n) \, \Delta x\), where \(f(x, y) = -3y\). Here, \(y_0 = 1.5\) and \(x_0 = 0\).

04

First Step (x = 0.2)

For the first step, \(x_1 = x_0 + \Delta x = 0.2\) and \(y' = -3y\). Calculate \(y_1\): \(y_1 = y_0 + (-3y_0) \, (0.2) = 1.5 + (-3 \, 1.5) \, 0.2 = 1.5 - 0.9 = 0.6\).

05

Second Step (x = 0.4)

Now, \(x_2 = x_1 + \Delta x = 0.4\) and calculate \(y_2\): \(y_2 = y_1 + (-3y_1) \, (0.2) = 0.6 + (-3 \, 0.6) \, 0.2 = 0.6 - 0.36 = 0.24\).

06

Third Step (x = 0.6)

Next, \(x_3 = x_2 + \Delta x = 0.6\) and calculate \(y_3\): \(y_3 = y_2 + (-3y_2) \, (0.2) = 0.24 + (-3 \, 0.24) \, 0.2 = 0.24 - 0.144 = 0.096\).

07

Fourth Step (x = 0.8)

Then, \(x_4 = x_3 + \Delta x = 0.8\) and calculate \(y_4\): \(y_4 = y_3 + (-3y_3) \, (0.2) = 0.096 + (-3 \, 0.096) \, 0.2 = 0.096 - 0.0576 = 0.0384\).

08

Fifth Step (x = 1.0)

Lastly, \(x_5 = x_4 + \Delta x = 1.0\) and calculate \(y_5\): \(y_5 = y_4 + (-3y_4) \, (0.2) = 0.0384 + (-3 \, 0.0384) \, 0.2 = 0.0384 - 0.02304 = 0.01536\).

09

Compare Estimates to Actual Solution

Using the actual solution \(y = 1.5 e^{-3x}\), compute \(y\) at the points calculated using Euler's method and compare them. For instance, at \(x = 0.2\), the actual solution is \(y = 1.5 e^{-0.6}\).

10

Verify y = 1.5e^{-3x} is a Solution

To verify \(y = 1.5 e^{-3x}\) is a solution to \(dy/dx = -3y\), compute its derivative: \(dy/dx = -3 \, (1.5 e^{-3x}) = -3y\). This matches the given differential equation. Additionally, when \(x=0, y = 1.5 e^{0} = 1.5\), which satisfies the initial condition \(y(0) = 1.5\).

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Key Concepts

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

Euler's Method

Euler's Method is a straightforward technique for numerically solving differential equations. It approximates the solution by taking small steps along the slope of the function.
Here’s how you can use Euler's Method for solving a differential equation like For each step, calculate the next value using the following formula: This method converts the problem into a series of small linear segments.
Although not exact, Euler's Method provides a reasonably accurate estimate, especially with smaller steps

Slope Field

A slope field (or direction field) is a visual representation of a differential equation’s solutions at given points in the To sketch a slope field for Calculate the slope at various points, e.g., for the point Draw tiny line segments through these points representing the direction of the slope.
Together, these segments form a ‘field’ showing possible directions of solution curves.
Slope fields are a great way to see the general behavior of solutions without actually solving the equation analytically.

Initial Conditions

Initial conditions specify the value of the solution at a particular point to uniquely determine a solution For instance, in the exercise, we are given the initial condition Initial conditions are essential because, without them, we can’t determine the specific solution curve out of the many potential solutions that fit the differential equation.

Solution Curve

The solution curve is the actual trajectory that follows the slope field, starting from the initial condition.
It’s the precise path that the function In our exercise, the solution curve for This curve helps visualize the behavior of the solution over time and compare it with numerical methods like Euler's

Analytical Solution

An analytical solution to a differential equation is a closed-form expression representing the exact solution.
For the differential equation This function exactly satisfies the given differential equation and initial condition.
Analytical solutions provide deeper insights and more accurate results compared to numerical methods, but they aren’t always easy or possible to obtain. Understanding all these concepts helps to more effectively solve differential equations and apply methods like Euler’s for numerical estimation.