Answer to Question #122589 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' = x2 + y2, y(0) = 3; y(0.5)

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

Answer 1

$\text{Euler's method} ~~ y_{n+1}= y_n+hf(x_n,y_n)\\[1 em] \text{First } h=0.1~~~~~~~~~~~~~ m=f(x_n,y_n)\\[1 em] ~~x_0=0 ~~~,~~~y_0=1.00~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~m_0=1.000\\[1 em] x_1=0.1~~~~y_1=y_0+m_0h=1+1(0.1)=1.1~~~~~~~~~m_1=1.22 \\[1 em] x_2=0.2~~~~y_2=y_1+m_1h=1.1+1.22(0.1)=1.22~~~~m_2=1.53 \\[1 em] x_3=0.3~~~~y_3=y_2+m_2h=1.373~~~~~~~~~~~~~ ~~~~~~~~~~~~~~m_3=1.975 \\[1 em] x_4=0.4~~~~y_4=y_3+m_3h=1.573~~~~~~~~~~~~~~~~~~~~~~~~~~~m_4=2.634 \\[1 em] x_5=0.5~~~~y_5=y_4+m_4h=1.8371 \\[1 em] \text{Second } h=0.05\\[1 em] ~~x_0=0 ~~~,~~~y_0=1.00~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~m_0=1.000\\[1 em] x_1=0.05~~,~~y_1=y_0+m_0h=1+1(0.05)=1.05~~~,~~~~~~m_1=1.105 \\[1 em] ~~~~x_2=0.1~~~~~~y_2=1.1053~~~~~~~~~m_2=1.2316 \\[1 em] ~~~~x_3=0.15~~~~~y_3=1.1668~~~~~~~~~m_3=2.8047 \\[1 em] ~~~~x_4=0.2~~~~~~y_4=1.2360~~~~~~~~~m_4=1.5677 \\[1 em] ~~~~x_5=0.25~~~~~y_5=1.3144~~~~~~~~~m_5=1.7901\\[1 em] ~~~~x_6=0.3~~~~~~y_6=1.4039~~~~~~~~~m_6=2.0610\\[1 em] ~~~~x_7=0.35~~~~y_7=1.5070~~~~~~~~~m_7=2.3934\\[1 em] ~~~~x_8=0.4~~~~~~y_8=1.6267~~~~~~~~~m_8=2.8060\\[1 em] ~~~~x_9=0.45~~~~y_9=1.7670~~~~~~~~~m_9=3.3248\\[1 em] ~~~~x_{10}=0.5~~~~y_{10}=1.9332\\[1 em] \therefore y(0.05)=1.8371 ~~~\text{for}~~h=0.1\\[1 em] ~~, y(0.05)=1.9332 ~~~\text{for}~~h=0.05\\[1 em]$