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Math  /  Calculus

QuestionUse Euler's Method with h=0.1\mathrm{h}=0.1 to approximate the solution to the following initial value problem on the interval 1≤x≤21 \leq x \leq 2. Compare these approximations with the actual solution y=−1xy=-\frac{1}{x} by graphing the polygonal-line approximation and the actual solution on the same coordinate system. y′=1x2−yx−y2,y(1)=−1y^{\prime}=\frac{1}{x^{2}}-\frac{y}{x}-y^{2}, y(1)=-1
Complete the table. \begin{tabular}{|c|c|c|} \hline n\boldsymbol{n} & xn\mathbf{x}_{\mathbf{n}} & \begin{tabular}{c} Euler's Method \\ approximate solution \end{tabular} \\ \hline 0 & 1 & -1 \\ \hline 1 & 1.1 & -0.90000 \\ \hline 2 & 1.2 & -0.81654 \\ \hline 3 & 1.3 & -0.74572 \\ \hline 4 & 1.4 & -0.68480 \\ \hline 5 & 1.5 & -0.63176 \\ \hline 6 & 1.6 & -0.58511 \\ \hline 7 & 1.7 & -0.54371 \\ \hline 8 & 1.8 & -0.50669 \\ \hline 9 & 1.9 & -0.47335 \\ \hline 10 & 2.0 & 0.44314 \\ \hline \end{tabular} (Round to five decimal places as needed.)

Studdy Solution

STEP 1

1. We are using Euler's Method to approximate the solution of a differential equation.
2. The step size h\mathrm{h} is given as 0.10.1.
3. The initial condition is y(1)=−1y(1) = -1.
4. The differential equation is y′=1x2−yx−y2y' = \frac{1}{x^2} - \frac{y}{x} - y^2.
5. The actual solution is y=−1xy = -\frac{1}{x}.

STEP 2

1. Understand Euler's Method.
2. Apply Euler's Method iteratively to approximate the solution.
3. Compare the Euler's Method approximation with the actual solution.
4. Graph the approximations and the actual solution.

STEP 3

Euler's Method is a numerical technique for solving ordinary differential equations (ODEs) with a given initial value. It uses the formula:
yn+1=yn+h⋅f(xn,yn) y_{n+1} = y_n + h \cdot f(x_n, y_n)
where f(x,y)=1x2−yx−y2 f(x, y) = \frac{1}{x^2} - \frac{y}{x} - y^2 is the derivative given in the problem.

STEP 4

Apply Euler's Method iteratively:
- Start at n=0 n = 0 with x0=1 x_0 = 1 and y0=−1 y_0 = -1 . - For each step, calculate yn+1=yn+0.1⋅f(xn,yn) y_{n+1} = y_n + 0.1 \cdot f(x_n, y_n) . - Update xn+1=xn+0.1 x_{n+1} = x_n + 0.1 .
Let's calculate each step:
- n=0 n = 0 , x0=1 x_0 = 1 , y0=−1 y_0 = -1 y1=y0+0.1⋅(112−−11−(−1)2)=−1+0.1⋅(1+1−1)=−0.9 y_1 = y_0 + 0.1 \cdot \left(\frac{1}{1^2} - \frac{-1}{1} - (-1)^2\right) = -1 + 0.1 \cdot (1 + 1 - 1) = -0.9
- n=1 n = 1 , x1=1.1 x_1 = 1.1 , y1=−0.9 y_1 = -0.9 y2=y1+0.1⋅(11.12−−0.91.1−(−0.9)2)≈−0.81654 y_2 = y_1 + 0.1 \cdot \left(\frac{1}{1.1^2} - \frac{-0.9}{1.1} - (-0.9)^2\right) \approx -0.81654
- Continue this process for n=2 n = 2 to n=10 n = 10 .

STEP 5

Compare the Euler's Method approximation with the actual solution y=−1x y = -\frac{1}{x} at each xn x_n .
- Calculate the actual solution values at each xn x_n . - Compare these values with the approximations obtained from Euler's Method.

STEP 6

Graph the approximations and the actual solution:
- Plot the points (xn,yn)(x_n, y_n) from Euler's Method. - Plot the curve y=−1x y = -\frac{1}{x} over the interval 1≤x≤2 1 \leq x \leq 2 .
The table is already completed with approximations using Euler's Method. The comparison and graphing are conceptual steps that would be done using graphing software or manually plotting on graph paper.

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