Answer to Question #126292 in Quantitative Methods for jse

Question #126292
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' = x^2 + y^2, y(0) = 3; y(0.5)

y(0.5) ≈ _______ (h = 0.1)
y(0.5) ≈ _______ (h = 0.05)
1
Expert's answer
2020-07-16T19:20:08-0400

Solution:

The iterative formulas for Euler's method are "x_{n+1}=x_n+h," "y_{n+1}=y_n+h\\times f(x_n, y_n)" , where "y'=f(x,y)" , "y(x_0)=y_0"

In this case "f(x,y)=x^2+y^2" , "x_0=0," "y_0=3"

I) Solution for h=0.1

"x_1=x_0+h=0+0.1=0.1"

"y_1=y_0+h\\times f(x_0, y_0)=3+0.1\\times (0^2+3^2)=3.9"

Analogically, searching for the next points of the solution - "x_2, y_2, x_3, y_3, x_4, y_4, x_5, y_5," until reaching the desired value y(0.5)

"x_2=x_1+h=0.1+0.1=0.2"

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

"x_3=x_2+h=0.3"

"y_3=y_3+h\\times f(x_3, y_3)" "=5.422+0.1\\times (0.2^2+5.422^2)=8.3658"

"x_4=x_3+h=0.4"

"y_4=y_4+h\\times f(x_4, y_4)" "=8.3658+0.1\\times (0.3^2+8.3658^2)=15.3735"

"x_5=x_4+h=0.5"

"y_5=y_4+h\\times f(x_4, y_4)" "=15.3735+0.1\\times (0.4^2+15.3735^2)=39.0239"

So the answer for h=0.1 is "y(0.5)=39.0239"

II)Solution for h=0.05

"x_1=x_0+h=0+0.05=0.05"

"y_1=y_0+h\\times f(x_0, y_0)" "=3+0.05\\times (0^2+3^2)=3.45"

"x_2=x_1+h=0.1"

"y_2=y_1+h\\times f(x_1, y_1)" "=3.45+0.05\\times (0.05^2+3.45^2)=4.0453"

"x_3=x_2+h=0.15"

"y_3=y_2+h\\times f(x_2, y_2)" "=4.0453+0.05\\times (0.1^2+4.0453^2)=4.8640"

"x_4=x_3+h=0.2"

"y_4=y_3+h\\times f(x_3, y_3)" "=4.8640+0.05\\times (0.15^2+4.8640^2)=6.0480"

"x_5=x_4+h=0.25"

"y_5=y_4+h\\times f(x_4, y_4)" "=6.0480+0.05\\times (0.2^2+6.0480^2)=7.8789"

"x_6=x_5+h=0.3"

"y_6=y_5+h\\times f(x_5, y_5)" "=7.8789+0.05\\times (0.25^2+7.8789^2)=10.9858"

"x_7=x_6+h=0.35"

"y_7=y_6+h\\times f(x_6, y_6)" "=10.9858+0.05\\times (0.3^2+10.9858^2)=17.0248"

"x_8=x_7+h=0.4"

"y_8=y_7+h\\times f(x_7, y_7)" "=17.0248+0.05\\times (0.35^2+17.0248^2)=31.5231"

"x_9=x_8+h=0.45"

"y_9=y_8+h\\times f(x_8, y_8)" "=31.5231+0.05\\times (0.4^2+31.5231^2)=81.2163"

"x_{10}=x_9+h=0.5"

"y_{10}=y_9+h\\times f(x_9, y_9)" "=81.2163+0.05\\times (0.45^2+81.2163^2)=411.0307"

So the answer for h=0.05 is "y(0.5)=411.0307"

Answer: "y(0.5)\\approx 39.0239 (h=0.1)"

"y(0.5) \\approx 411.0307 (h=0.05)"

Results of all the calculations are summarized in a tabular form (screenshot attached)

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