y' = x-3
P0 = (1,4)
∆y = y'(x0)*∆x = (-2)(0.1) = -0.2
P1 = (1.1,3.8)
∆y = y'(x1)*∆x = (-1.9)(0.1) = -0.19
P2 = (1.2,3.61)
∆y = y'(x2)*∆x = (-1.8)(0.1) = -0.18
P3 = (1.3,3.43)
Read the fine article at wikipedia, which shows a geometric interpretation of the method.
Hello! Just needed help with understanding this specific question. We never really wet over Euler's method so I'm not sure how to go about it.
"Use Euler's method in order to solve the initial value problem below.
dy/dx = x-3 and y=4 when x=1
Use Euler's method with increment's of (triangle symbol)x=0.1 to approximate the value of y when x=1.3
A. 3.48
B. 3.68
C. 3.5
D. 3.43"
Thank you so much for your help.
1 answer
1 answer