Consider the following. (A computer algebra system is recommended. Round your an
ID: 3195608 • Letter: C
Question
Consider the following. (A computer algebra system is recommended. Round your answers to four decimal places.) y, 5 + t-y, y(0) = 1 (a) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h 0.1. y(0.2) v(0.3) y(0.4) = (b) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.05 y(0.1) y(0.3) y(0.4) = (c) Find approximate values of the solution of the given initial value problem at t = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h-0.025 y(0.1) = y(0.2) (0.3) y(0.4) = (d) Find the solution y = (t) of the given problem y(t) = Evaluate g(t) at t = 0.1, 0.2, 0.3, and 0.4. (Round your answers to five decimal places.) y(0.1) = y(0.2) y(0.3) y(0.4) =Explanation / Answer
The Euler's formula is :
yn+1= yn+hf(tn,yn)
Given : y' = 5+t-y y(0)= 1
(a)
y1 = y(0.1) = y0 +hf(y0) = 1 +0.1(4) = 1.4
y2 = y(0.2)= y1+hf(0.1,1.4) = 1.4 +0.1(3.7)= 1.77
y3 = y(0.3)= y2+hf(0.2,1.77)= 1.77+0.1(3.43)= 2.113
y4= y(0.4)= y3+hf(0.3,2.113) = 3.187
(b)
y1 = y(0.05)= y0+hf(0,1)= 1+(0.05)(4)= 1.2
y2= y(0.1)= y1+hf(0.05,1.2)= 1.2+(0.05)(3.85)= 1.3925
y3 = y(0.15)= y2+hf(0.1,1.3925) = 1.3925 +(0.05)(3.7075) = 1.5779
y4 = y(0.2)= y3+hf(0.15,1.5779) = 1.7565
y5 = y(0.25)= y4+hf(0.2,1.7565)= 1.9287
y6 = y(0.3)= y5+hf(0.25,1.9287)= 2.0948
y7 = y(0.35)= y6+hf(0.3,2.0987)= 2.2588
y8 = y(0.4)= y7+hf(0.35,2.2588)= 2.4134
(c)
y(0.025)= 1+(0.025)(4) = 1.1
y(0.05)= 1.1+ (0.025(3.925)= 1.1981
y(0.075) = 1.1981+ (0.025)(3.8519)= 1.2944
y(0.1)= 1.2944+ (0.025)(3.7806)= 1.3889
y(0.125)= 1.3889+ (0.025)(3.7111)= 1.4817
y(0.15)= 1.4817+ 0.025(3.6683)= 1.5734
y(0.175)= 1.5734+ 0.025(3.6016) =1.6634
y(0.2)= 1.6634+0.025(3.5366)= 1.7518
And so on..
(d)
y' = 5+t-y
Its general solution is :
y(t)= c1e-t+ t+ 4
The initial condition is : y(0)= 1
c1 +4 = 1
c1 = -3
Hence
y(t)= -3e-t+t+4