Use Euler’s method to approximate y(0.2), where y(x) is the solution of the initial-value problem y”-(2x + 1)y =1, y(0)=3, y'(0) ?=1

Use Euler’s method to approximate y(0.2), where y(x) is the solution of the initial-value problem y”-(2x + 1)y =1, y(0)=3, y'(0) ?=1. First use one step with h =0.2 and then repeat the calculations using two steps with h =0.1.

we shall rewrite this equation as folllows:

$\f[y''=1+(2x+1)y\f]$

we then let

$\f[u=y'\f]$

thus

$\f[u'=1+(2x+1)y\f]$

the recursive euler’s method is

$\f[y_{n+1}=y_{n}+hu_{n}\f]$

$\f[u_{n+1}=u_{n}+hf(x_{n},y_{n},u_{n})\f]$

starting with h=0.2;

$\f[u_{0}=1\f]$

thus

$\f[y_{1}=y_{0}+hu_{0}\f]$

$\f[y_{1}=y(0.2)=3+0.2(1)=3.2\f]$

then h=0.1;

$\f[y_{1}=y_{0}+hu_{0}\f]$

$\f[y_{1}=3+(0.1*1)=3.1\f]$

also

$\f[u_{1}=u_{0}+hf(x_{0},y_{0},u_{0})\f]$

$\f[u_{1}=1+0.1[1+(2*0+1)3]=1.4\f]$

then

$\f[y_{2}=y_{1}+hu_{1}\f]$

$\f[y_{2}=y(0.2)=3.1+(0.1*1.4)=3.24\f]$

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Use Euler’s method to approximate y(0.2), where y(x) is the solution of the initial-value problem y”-(2x + 1)y =1, y(0)=3, y'(0) ?=1

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