Answer to Question #126292 in Quantitative Methods for jse

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

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

Answer 1

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)