y'' + 4y' + 4y = 0 ---- y(0) = 1, y'(0) = 5
Find the exact solution of the differential equation.
Use the exact solution and Euler's Method to compute Euler's Approximation for time t = 0 to t = 5 using a step h=0.05. Plot Euler's & Exact vs. t and plot Error vs. t. Then, answer the following question:
Is the error really increasing? That is, is Euler's method becoming less accurate as t increases?
x1 = y
x2 = y'
x1' = y'
x2' = y'' = -4y' - 4y = -4x2 - 41
Euler's Method Equations assuming the substitutions made above:
x1(n+1) = x1(n) + hx/2(n)
x2(n+1) = x2(n) + h*(y''(n))
The Attempt at a Solution
First, I solved the differential equation. Writing the characteristic equation r2 + 4r + 4 = 0, I solved for there to be a repeated root at r = -2. Therefore, the general solution is:
y= c1e-2t + c2te-2t
Taking a derivative to use the initial conditions:
y' = -c1/2 * e-2t + c2te-2t/2 + c2e-2t
Applying the initla conditions, I find c1 = 1 and c2 = 11/2
Therefore, the solution becomes:
e-2t + 11t/2 * e-2t
I then created an excel spreadsheet which calculate and then plots the exact solution vs. the eulerapproximation, giving me the graph:
then the error graph looks as follows:
So around time t = 3, it makes sense that the error would be about 0 due to how Euler approximations work. Since Euler Approximations use the tangent lines of the functions at a certain point to approximate the shape of the graph, the more and more linear the solution gets (such as from t --> infinity in this solution), the easier it is to approximate the solution because the slope of a horizontal line is 0. However, I have absolutely no idea what could be going on here that would cause the error to start increasing seemingly linearly after time t = 3 where the solution and the euler approximation is practically horizontal. I cannot fathom that a repeated root would cause this error. Does anyone have any idea what could be causing this?
Thanks in advance.