Draw a directional field to get an idea of what the solution is:
t,y=var('t,y')
df = 3*cos(t)-2*y
v=plot_slope_field(df,(t,0,3),(y,-3,3))
plot(v)
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#4a. 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
# differential equation
f(t,y) = 3*cos(t)-2*y
# initial value
t = 0
y = 0
# step size
h = 0.1
# Euler's method y_n+1 = y_n + f_n*h
for j in range(4) :
y = y + f(t,y)*h
t = t + h
print t.n(digits=2), y.n(digits=4)
0.10 0.3000 0.20 0.5385 0.30 0.7248 0.40 0.8665 0.10 0.3000 0.20 0.5385 0.30 0.7248 0.40 0.8665 |
#4b. Repeat part (a) with h = 0.05. Compare the results with those found in (a).
f(t,y) = 3*cos(t)-2*y
t = 0
y = 0
h = 0.05
for i in range(4) :
for j in range(2) :
y = y + f(t,y)*h
t = t + h
print t.n(digits=2), y.n(digits=5)
0.10 0.28481 0.20 0.51334 0.30 0.69345 0.40 0.83157 0.10 0.28481 0.20 0.51334 0.30 0.69345 0.40 0.83157 |
#4c. Repeat part (a) with h = 0.025. Compare the results with those found in (a) and (b).
f(t,y) = 3*cos(t)-2*y
t = 0
y = 0
h = 0.025
for i in range(4) :
for j in range(4) :
y = y + f(t,y)*h
t = t + h
print t.n(digits=2), y.n(digits=5)
0.10 0.27792 0.20 0.50181 0.30 0.67895 0.40 0.81530 0.10 0.27792 0.20 0.50181 0.30 0.67895 0.40 0.81530 |
#4d. Find the solution of the given problem and evaluate at t = 0.1, 0.2, 0.3, and 0.4. Compare these values with the results of (a), (b), and (c).
x = var('t')
y = function('y', t)
a = desolve(diff(y,t)-3*cos(t)+2*y, y, ics=[0,0])
a.expand()
6/5*cos(t) - 6/5*e^(-2*t) + 3/5*sin(t) 6/5*cos(t) - 6/5*e^(-2*t) + 3/5*sin(t) |
t = 0.1
h = 0.1
for j in range(4) :
a(t)
t = t + h
0.271428144628150 0.490897436643760 0.665141947634699 0.799729441247987 0.271428144628150 0.490897436643760 0.665141947634699 0.799729441247987 |
d = contour_plot(C, (x,-2, 2), (y,-2,2),fill = False, linewidths = 5, contours = 30, cmap = 'hsv')
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