Math Problem Statement

Use Euler’s method with stepwise 0.1 to estimate 𝑦(0.5), where 𝑦(π‘₯) is the solution of the initial value problem 𝑦′ = 𝑦 + π‘₯𝑦, 𝑦(0) = 1

Solution

Euler's method is a numerical technique used to approximate the solution of a differential equation. In this case, we will use Euler's method with a step size of h=0.1h = 0.1 to estimate y(0.5)y(0.5) for the differential equation yβ€²=y+xyy' = y + xy with the initial condition y(0)=1y(0) = 1.

Step-by-Step Procedure

  1. Initial Setup: yβ€²(x)=y+xyy'(x) = y + xy y(0)=1y(0) = 1 h=0.1h = 0.1

  2. Euler's Method Formula: yn+1=yn+hβ‹…f(xn,yn)y_{n+1} = y_n + h \cdot f(x_n, y_n) Where f(x,y)=y+xyf(x, y) = y + xy.

  3. Iterative Calculation:

    • y0=1y_0 = 1
    • For each step, calculate yn+1y_{n+1} using the formula above.

Let's calculate y(x)y(x) at each step from x=0x = 0 to x=0.5x = 0.5.

Step-by-Step Calculation:

Step 1: x=0x = 0 to x=0.1x = 0.1 x0=0,y0=1x_0 = 0, \quad y_0 = 1 f(x0,y0)=y0+x0y0=1+0β‹…1=1f(x_0, y_0) = y_0 + x_0 y_0 = 1 + 0 \cdot 1 = 1 y1=y0+hβ‹…f(x0,y0)=1+0.1β‹…1=1.1y_1 = y_0 + h \cdot f(x_0, y_0) = 1 + 0.1 \cdot 1 = 1.1

Step 2: x=0.1x = 0.1 to x=0.2x = 0.2 x1=0.1,y1=1.1x_1 = 0.1, \quad y_1 = 1.1 f(x1,y1)=y1+x1y1=1.1+0.1β‹…1.1=1.21f(x_1, y_1) = y_1 + x_1 y_1 = 1.1 + 0.1 \cdot 1.1 = 1.21 y2=y1+hβ‹…f(x1,y1)=1.1+0.1β‹…1.21=1.221y_2 = y_1 + h \cdot f(x_1, y_1) = 1.1 + 0.1 \cdot 1.21 = 1.221

Step 3: x=0.2x = 0.2 to x=0.3x = 0.3 x2=0.2,y2=1.221x_2 = 0.2, \quad y_2 = 1.221 f(x2,y2)=y2+x2y2=1.221+0.2β‹…1.221=1.4652f(x_2, y_2) = y_2 + x_2 y_2 = 1.221 + 0.2 \cdot 1.221 = 1.4652 y3=y2+hβ‹…f(x2,y2)=1.221+0.1β‹…1.4652=1.36752y_3 = y_2 + h \cdot f(x_2, y_2) = 1.221 + 0.1 \cdot 1.4652 = 1.36752

Step 4: x=0.3x = 0.3 to x=0.4x = 0.4 x3=0.3,y3=1.36752x_3 = 0.3, \quad y_3 = 1.36752 f(x3,y3)=y3+x3y3=1.36752+0.3β‹…1.36752=1.777776f(x_3, y_3) = y_3 + x_3 y_3 = 1.36752 + 0.3 \cdot 1.36752 = 1.777776 y4=y3+hβ‹…f(x3,y3)=1.36752+0.1β‹…1.777776=1.5452976y_4 = y_3 + h \cdot f(x_3, y_3) = 1.36752 + 0.1 \cdot 1.777776 = 1.5452976

Step 5: x=0.4x = 0.4 to x=0.5x = 0.5 x4=0.4,y4=1.5452976x_4 = 0.4, \quad y_4 = 1.5452976 f(x4,y4)=y4+x4y4=1.5452976+0.4β‹…1.5452976=2.16341664f(x_4, y_4) = y_4 + x_4 y_4 = 1.5452976 + 0.4 \cdot 1.5452976 = 2.16341664 y5=y4+hβ‹…f(x4,y4)=1.5452976+0.1β‹…2.16341664=1.761639264y_5 = y_4 + h \cdot f(x_4, y_4) = 1.5452976 + 0.1 \cdot 2.16341664 = 1.761639264

Final Result

The estimated value of y(0.5)y(0.5) using Euler's method with a step size of 0.1 is approximately: y(0.5)β‰ˆ1.7616y(0.5) \approx 1.7616

Would you like more details or have any questions?

Suggested Questions to Ask:

  1. How does the step size in Euler's method affect the accuracy of the solution?
  2. What are some limitations of Euler's method?
  3. Can you explain the concept of numerical stability in relation to Euler's method?
  4. How can we improve the accuracy of the solution obtained using Euler's method?
  5. What are some alternative numerical methods to solve differential equations?

Tip:

When using numerical methods like Euler's method, smaller step sizes typically yield more accurate results, but at the cost of increased computational effort. Balancing accuracy and efficiency is key in numerical analysis.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Numerical Methods
Euler's Method

Formulas

Euler's Method: y_{n+1} = y_n + h \cdot f(x_n, y_n)
Differential Equation: y' = y + xy

Theorems

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Suitable Grade Level

College Level