7.3. EULER’S METHOD
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2. We have seen that the error in approximating the solution to an initial value problem is
proportional to ∆t. That is, if E ∆t is the Euler’s method approximation to the solution
to an initial value problem at t, then
y(t) − E ∆t ≈ K∆t
for some constant of proportionality K.
In this problem, we will see how to use this fact to improve our estimates, using an idea
called accelerated convergence.
(a) We will create a new approximation by assuming the error is exactly proportional to ∆t, according to the formula
y(t) − E ∆t = K∆t.
Using our earlier results from the initial value problem dy/dt = y and y(0) = 1
with ∆t = 0.2 and ∆t = 0.1, we have
y(1) − 2.4883 = 0.2K
y(1) − 2.5937 = 0.1K .
This is a system of two linear equations in the unknowns y(1) and K. Solve
this system to find a new approximation for y(1). (You may remember that the
exact value is y(1) = e = 2.71828 . . . .)
(b) Use the other data, E 0.05 = 2.6533 and E 0.025 = 2.6851 to do similar work as
in (a) to obtain another approximation. Which gives the better approximation?
Why do you think this is?
(c) Let’s now study the initial value problem
dy
dt
= t − y, y(0) = 0.
Approximate y(0.3) by applying Euler’s method to find approximations E 0.1
and E 0.05 . Now use the idea of accelerated convergence to obtain a better
approximation. (For the sake of comparison, you want to note that the actual
value is y(0.3) = 0.0408.)
3. In this problem, we’ll modify Euler’s method to obtain better approximations to
solutions of initial value problems. This method is called the Improved Euler’s method.
In Euler’s method, we walk across an interval of width ∆t using the slope obtained from
the differential equation at the left endpoint of the interval. Of course, the slope of the
417
2. We have seen that the error in approximating the solution to an initial value problem is
proportional to ∆t. That is, if E ∆t is the Euler’s method approximation to the solution
to an initial value problem at t, then
y(t) − E ∆t ≈ K∆t
for some constant of proportionality K.
In this problem, we will see how to use this fact to improve our estimates, using an idea
called accelerated convergence.
(a) We will create a new approximation by assuming the error is exactly proportional to ∆t, according to the formula
y(t) − E ∆t = K∆t.
Using our earlier results from the initial value problem dy/dt = y and y(0) = 1
with ∆t = 0.2 and ∆t = 0.1, we have
y(1) − 2.4883 = 0.2K
y(1) − 2.5937 = 0.1K .
This is a system of two linear equations in the unknowns y(1) and K. Solve
this system to find a new approximation for y(1). (You may remember that the
exact value is y(1) = e = 2.71828 . . . .)
(b) Use the other data, E 0.05 = 2.6533 and E 0.025 = 2.6851 to do similar work as
in (a) to obtain another approximation. Which gives the better approximation?
Why do you think this is?
(c) Let’s now study the initial value problem
dy
dt
= t − y, y(0) = 0.
Approximate y(0.3) by applying Euler’s method to find approximations E 0.1
and E 0.05 . Now use the idea of accelerated convergence to obtain a better
approximation. (For the sake of comparison, you want to note that the actual
value is y(0.3) = 0.0408.)
3. In this problem, we’ll modify Euler’s method to obtain better approximations to
solutions of initial value problems. This method is called the Improved Euler’s method.
In Euler’s method, we walk across an interval of width ∆t using the slope obtained from
the differential equation at the left endpoint of the interval. Of course, the slope of the