Math Problem Statement

Do the Euler’s Numerical Method

Do the Euler’s Numerical Method

Solution

To solve the given differential equation using Euler’s numerical method with step size h=0.1h = 0.1, we proceed as follows:

The given differential equation is: dydx+2xy=x,y(0)=2\frac{dy}{dx} + 2xy = x, \quad y(0) = 2

Rewriting it in standard form for Euler's method: dydx=x2xy\frac{dy}{dx} = x - 2xy

Define: f(x,y)=x2xyf(x, y) = x - 2xy

Step-by-step process:

Initial conditions:

x0=0,y0=2x_0 = 0, \quad y_0 = 2

We calculate yy iteratively at x=0.1,0.2,,0.6x = 0.1, 0.2, \dots, 0.6.

Iteration formula:

yn+1=yn+hf(xn,yn)y_{n+1} = y_n + h \cdot f(x_n, y_n)

Here, h=0.1h = 0.1.

Iterations:

  1. At x0=0,y0=2x_0 = 0, y_0 = 2: f(0,2)=02(0)(2)=0f(0, 2) = 0 - 2(0)(2) = 0 y1=y0+hf(0,2)=2+0.10=2y_1 = y_0 + h \cdot f(0, 2) = 2 + 0.1 \cdot 0 = 2

  2. At x1=0.1,y1=2x_1 = 0.1, y_1 = 2: f(0.1,2)=0.12(0.1)(2)=0.10.4=0.3f(0.1, 2) = 0.1 - 2(0.1)(2) = 0.1 - 0.4 = -0.3 y2=y1+hf(0.1,2)=2+0.1(0.3)=20.03=1.97y_2 = y_1 + h \cdot f(0.1, 2) = 2 + 0.1 \cdot (-0.3) = 2 - 0.03 = 1.97

  3. At x2=0.2,y2=1.97x_2 = 0.2, y_2 = 1.97: f(0.2,1.97)=0.22(0.2)(1.97)=0.20.788=0.588f(0.2, 1.97) = 0.2 - 2(0.2)(1.97) = 0.2 - 0.788 = -0.588 y3=y2+hf(0.2,1.97)=1.97+0.1(0.588)=1.970.0588=1.9112y_3 = y_2 + h \cdot f(0.2, 1.97) = 1.97 + 0.1 \cdot (-0.588) = 1.97 - 0.0588 = 1.9112

  4. At x3=0.3,y3=1.9112x_3 = 0.3, y_3 = 1.9112: f(0.3,1.9112)=0.32(0.3)(1.9112)=0.31.14672=0.84672f(0.3, 1.9112) = 0.3 - 2(0.3)(1.9112) = 0.3 - 1.14672 = -0.84672 y4=y3+hf(0.3,1.9112)=1.9112+0.1(0.84672)=1.91120.084672=1.826528y_4 = y_3 + h \cdot f(0.3, 1.9112) = 1.9112 + 0.1 \cdot (-0.84672) = 1.9112 - 0.084672 = 1.826528

  5. At x4=0.4,y4=1.826528x_4 = 0.4, y_4 = 1.826528: f(0.4,1.826528)=0.42(0.4)(1.826528)=0.41.4612224=1.0612224f(0.4, 1.826528) = 0.4 - 2(0.4)(1.826528) = 0.4 - 1.4612224 = -1.0612224 y5=y4+hf(0.4,1.826528)=1.826528+0.1(1.0612224)=1.8265280.10612224=1.72040576y_5 = y_4 + h \cdot f(0.4, 1.826528) = 1.826528 + 0.1 \cdot (-1.0612224) = 1.826528 - 0.10612224 = 1.72040576

  6. At x5=0.5,y5=1.72040576x_5 = 0.5, y_5 = 1.72040576: f(0.5,1.72040576)=0.52(0.5)(1.72040576)=0.51.72040576=1.22040576f(0.5, 1.72040576) = 0.5 - 2(0.5)(1.72040576) = 0.5 - 1.72040576 = -1.22040576 y6=y5+hf(0.5,1.72040576)=1.72040576+0.1(1.22040576)=1.720405760.122040576=1.598365184y_6 = y_5 + h \cdot f(0.5, 1.72040576) = 1.72040576 + 0.1 \cdot (-1.22040576) = 1.72040576 - 0.122040576 = 1.598365184

  7. At x6=0.6,y6=1.598365184x_6 = 0.6, y_6 = 1.598365184: f(0.6,1.598365184)=0.62(0.6)(1.598365184)=0.61.9180382208=1.3180382208f(0.6, 1.598365184) = 0.6 - 2(0.6)(1.598365184) = 0.6 - 1.9180382208 = -1.3180382208 y7=y6+hf(0.6,1.598365184)=1.598365184+0.1(1.3180382208)=1.5983651840.13180382208=1.46656136192y_7 = y_6 + h \cdot f(0.6, 1.598365184) = 1.598365184 + 0.1 \cdot (-1.3180382208) = 1.598365184 - 0.13180382208 = 1.46656136192

Final Result:

y(0.6)1.467(rounded to three decimal places)y(0.6) \approx 1.467 \, (\text{rounded to three decimal places})


Would you like a detailed comparison with other methods or visual representation of the solution?

Related Questions:

  1. How can this solution be checked using an exact or analytical solution?
  2. What is the error associated with using Euler's method in this case?
  3. How does the step size hh affect the accuracy of Euler's method?
  4. Can higher-order methods like the Runge-Kutta method improve the accuracy here?
  5. How does the behavior of the differential equation change for different initial conditions?

Tip: Euler's method is a simple numerical approach but can accumulate significant errors for large step sizes.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Numerical Methods
Euler's Method

Formulas

y_{n+1} = y_n + h * f(x_n, y_n)
f(x, y) = x - 2xy

Theorems

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Suitable Grade Level

Undergraduate