Solve the differential equation #y'+2y=3x+1# with initial conditions #y(1) = -1# using Euler approximation ?

1 Answer
Cesareo R.
Jan 29, 2017

See below.

Explanation:

Making the Euler differences we have

#(y_k-y_(k-1))/h=1+3x_(k-1)-2y_(k-1)# or

#y_k=(1-2h)y_(k-1)+h(1+3x_(k-1))#. Now beginning with #x_0=1,y_0=2# and knowing that #x_k = 1+kh, k=0,1,cdots,n# we can calculate the #y_k#. Here #n=floor((2-1)/h)#

Follow a comparison between the two Euler approximations and the exact solution of the differential equation which is

#y=1/4(3e^(2(1-x))+6x-1)#

The coarser approximation is for #h=0.2# follows the approximation for #h=0.1# and also was included an approximation for #h=0.02#

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