for the initial value problemy 2y 2 - e-4t y0

Question

For the initial value problem

y' + 2y = 2 - e^-4t, y(0) = 1

By using Euler's Method along with a step size of h = 0.1 to get approximate values of the solution at t = 0.1, 0.2, 0.3, 0.4, and 0.5. Compare them to the accurate values of the solution as such points.

Solution

It is a fairly simple linear differential equation thus we'll leave it to you to check that the solution as

y(t) = 1 + ½ e^-4t - ½ e^-2t

Thus as to use Euler's Method we first want to rewrite the differential equation in the form specified in (1).

y'= 2 - e^-4t-2y

From that we can notice that f (t, y ) = 2 - e^-4t  - 2y.  Also see that t = 0 and y₀ = 1.  We can here start doing many computations.

f_o = f(0,1) = 2 - e^-4(0)  - 2(1) = -1

y₁ = y₀ + h f₀ = 1 (0.1) (-1) = 0.9

Therefore, the approximation to the solution at t₁ = 0.1 is y₁ = 0.9.

At the next step we contain

f₁ = f(0.1,0.9) = 2 - e^-4(0.1)  - 2(0.9) = -0.470320046

y₂ = y₁ + h f₁ = 0.9 + (0.1) (-0.470320046) = 0.852967995

Therefore, the approximation to the solution at t2 = 0.2 is y2 = 0.852967995.

I'll leave this to you to verify the remainder of these calculations.

 f₂  =-0.155264954,     y₃  = 0.837441500

f₃  =0.023922788,        y₄  = 0.839833779

f₄  =0.1184359245,      y₅  = 0.851677371

Here's a rapid table which gives the approximations and also the exact value of the solutions at the specified points.

Time, tn	Approximation	Exact	Error
t0 = 0 t1 = 0.1 t2 = 0.2 t3 = 0.3 t4 = 0.4 t5 = 0.5	y0 =1 y1 =0.9 y2 =0.852967995 y3 =0.837441500 y4 =0.839833779 y5 =0.851677371	y(0) = 1 y(0.1) = 0.925794646 y(0.2) = 0.889504459 y(0.3) = 0.876191288 y(0.4) = 0.876283777 y(0.5) = 0.883727921	0 % 2.79 % 4.11 % 4.42 % 4.16 % 3.63 %

We've also comprised the error as a percentage. It's frequently easier to notice how well an approximation does whether you look at percentages. The formula for that is,

 Percent error = (|exact - approximate|/exact) - 100

We utilized absolute value in the numerator because we actually don't care at this point if the approximation is smaller or larger than the exact. We're merely interested in how close the two are.

The maximum error in the approximations from the previous illustration was 4.42 percent that isn't too bad, although also isn't all that great of an approximation. Thus, provided we aren't after very correct approximations such didn't do too badly. This type of error is commonly unacceptable in almost all actual applications though. Consequently, how can we get better approximations?

By using a tangent line recall that we are getting the approximations to approximate the value of the solution and which we are moving forward in time through steps of h. Therefore, if we need a more accurate approximation, so it seems like one manner to get a better approximation is to not move forward as much along with each step. Conversely, take smaller h's.