considering the initial value problem P'(t)= 0.4P(t), P(0)=1000, how would you compute the first four steps of Euler's method with a step size of t=1

Question

this is confusing me because I'm used to questions with a 'y' instead of a function inside the initial value problem.

Answer 1

P' is the slope of the tangent line.

Since (0, 1000) is a point on the curve, and since P'(t) = 0.4P(t) then we can calculate P'(0):

P'(0) = 0.4(1000) = 400

Creating a linear function using a slope of 400 and the point (0, 1000) gives us:

y - 1000 = 400(x - 0) or y = 400x + 1000 or, in this case, P(t) = 400t + 1000

Plugging in t = 1 (a step size of 1), we estimate the function value at t = 1 as:

P(1) = 400(1) + 1000 = 1400 so our first step yields the point (1, 1400).

At t = 1, P'(1) = 0.4P(1) = 0.4(1400) = 560 so the slope at t = 1 is 560. Making a linear function with a slope of 560 and a point (1, 1400) we get:

y - 1400 = 560(x - 1) or y = 560x + 840 or, in this case P(t) = 560t + 840

Plugging in t = 2 (step size of 1), we estimate the function value at t = 2 as:

P(2) = 560(2) + 840 = 1960 so 2nd step yields the point (2, 1960)

At t = 2, P'(2) = 0.4(1960) = 784 so the slope at t = 2 is 784. Making a linear function with a slope of 784 and a point (2, 1960) we get:

y - 1960 = 784(x - 2) or y = 784x + 392 or, in this case P(t) = 784t + 392

Plugging in t = 3 (step size of 1), we estimate the function value at t = 3 as:

P(3) = 784(3) + 392 = 2744 so 3rd step yields the point (3, 2744)

At t = 3, P'(3) = 0.4(2744) = 1097.6 so the slope at t = 3 is 1097.6. Making a linear function with a slope of 1097.6 and a point (3, 2744) we get:

y - 2744 = 1097.6(x - 3) or y = 1097.6x - 548.8 or, in this case P(t) = 1097.6t - 548.8

Plugging in t = 4 (step size of 1), we estimate the function value at t = 4 as:

P(4) = 1097.6(4) - 548.8 = 3841.6 so 4th step yields the point (4, 3841.6)