Answer to Question #125138 in Quantitative Methods for joseph se

Question

Use a numerical solver and Euler's method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05.

y' = x2 + y2, y(0) = 3; y(0.5)

y(0.5) ≈ _________ (h = 0.1)
y(0.5) ≈ _________(h = 0.05)

Answer 1

Euler's method explanation

We have an equation $y'=f(x,y)$, a point $(x_o,y_o)$ and a step value $h$. We determine a sequence of points $(x_1,y_1), (x_2,y_2) \dots (x_n,y_n)$, where $x_{i+1} = x_i + h$ and $y_{i+1} = y_i + h\cdot f(x_i,y_i)$.

In our task $f(x,y) = x^2+y^2$, point is $(0,3)$, and using current step value we want to find the point $(x_n, y_n)$, where $x_n = 0.5$.

Euler's method with step h = 0.1

Solution

$x_1 = x_o + h = 0 +0.1 = 0.1$

$y_1 = y_o + h \cdot f(x_o,y_o) = 3 + 0.1\cdot(0^2+3^2) = 3.9$

$x_2 = x_1 + h = 0.1 + 0.1 = 0.2$

$y_2 = y_1 + h \cdot f(x_1,y_1) = 3.9 + 0.1\cdot(0.1^2+3.9^2) = 5.422$

$x_3 = x_2 + h = 0.2 + 0.1 = 0.3$

$y_3 = y_2 + h \cdot f(x_2,y_2) = 5.422 + 0.1\cdot(0.2^2+ 5.422^2) \approx 8.3658$

$x_4 = x_3 + h = 0.3 + 0.1 = 0.4$

$y_4 = y_3 + h \cdot f(x_3,y_3) = 8.3658 + 0.1\cdot(0.3^2+ 8.3658^2) \approx 15.3735$

$x_5 = x_4 + h = 0.4 + 0.1 = 0.5$

$y_5 = y_4 + h \cdot f(x_4,y_4) = 15.3735 + 0.1\cdot(0.4^2+ 15.3735^2) \approx 39.0240$

Answer: y(0.5) ≈ 39.024

Euler's method with step h = 0.05

Solution

$x_1 = x_o + h = 0 + 0.05 = 0.05\\ y_1 = y_o + h \cdot f(x_o, y_o) = 3 + 0.05 \cdot (0^2 + 3^2) = 3.45\\ x_2 = x_1 + h = 0.05 + 0.05 = 0.1\\ y_2 = y_1 + h \cdot f(x_1, y_1) = 3.45 + 0.05 \cdot (0.05^2 + 3.45^2) \approx 4.0453\\ x_3 = x_2 + h = 0.1 + 0.05 = 0.15\\ y_3 = y_2 + h \cdot f(x_2, y_2) = 4.0453 + 0.05 \cdot (0.1^2 + 4.0453^2) \approx 4.864\\ x_4 = x_3 + h = 0.15 + 0.05 = 0.2\\ y_4 = y_3 + h \cdot f(x_3, y_3) = 4.864 + 0.05 \cdot (0.15^2 + 4.864^2) \approx 6.048\\ x_5 = x_4 + h = 0.2 + 0.05 = 0.25\\ y_5 = y_4 + h \cdot f(x_4, y_4) = 6.048 + 0.05 \cdot (0.2^2 + 6.048^2) \approx 7.8789\\ x_6 = x_5 + h = 0.25 + 0.05 = 0.3\\ y_6 = y_5 + h \cdot f(x_5, y_5) = 7.8789 + 0.05 \cdot (0.25^2 + 7.8789^2) \approx 10.9859\\ x_7 = x_6 + h = 0.3 + 0.05 = 0.35\\ y_7 = y_6 + h \cdot f(x_6, y_6) = 10.9859 + 0.05 \cdot (0.3^2 + 10.9859^2) \approx 17.0249\\ x_8 = x_7 + h = 0.35 + 0.05 = 0.4\\ y_8 = y_7 + h \cdot f(x_7, y_7) = 17.0249 + 0.05 \cdot (0.35^2 + 17.0249^2) \approx 31.5234\\ x_9 = x_8 + h = 0.4 + 0.05 = 0.45\\ y_9 = y_8 + h \cdot f(x_8, y_8) = 31.5234 + 0.05 \cdot (0.4^2 + 31.5234^2) \approx 81.2176\\ x_{10} = x_9 + h = 0.45 + 0.05 = 0.5\\ y_{10} = y_9 + h \cdot f(x_9, y_9) = 81.2176 + 0.05 \cdot (0.45^2 + 81.2176^2) \approx 411.0427\\$

Answer: y(0.5) ≈ 411.0427