- #1

evinda

Gold Member

MHB

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Hello! (Wave)

We consider the initial value problem:

$$x'(t)=-5x(t)-2y(t), t \in [0,1] \\ y'(t)=-2x(t)-100y(t), t \in [0,1] \\ x(0)=1, y(0)=1.$$

I want to solve the above problem using the forward Euler method, the trapezoid method and the backward Euler method and to represent in common graphs the corresponding values of $(x^n)^2+(y^n)^2$.

First of all, the formula for the approximation $y$ that we have using the backward Euler method is the following, right?

[m]y=inv(eye(2)-h*A)*y [/m]

where [m]A=[-5 -2;-2 -100][/m], right?

Then, I get the following graph for the approximations $(x^n)^2+(y^n)^2$.

View attachment 8113

The graphs are approximately the same, aren't they?

How can we deduce from this graph which method is better? (Thinking)

We consider the initial value problem:

$$x'(t)=-5x(t)-2y(t), t \in [0,1] \\ y'(t)=-2x(t)-100y(t), t \in [0,1] \\ x(0)=1, y(0)=1.$$

I want to solve the above problem using the forward Euler method, the trapezoid method and the backward Euler method and to represent in common graphs the corresponding values of $(x^n)^2+(y^n)^2$.

First of all, the formula for the approximation $y$ that we have using the backward Euler method is the following, right?

[m]y=inv(eye(2)-h*A)*y [/m]

where [m]A=[-5 -2;-2 -100][/m], right?

Then, I get the following graph for the approximations $(x^n)^2+(y^n)^2$.

View attachment 8113

The graphs are approximately the same, aren't they?

How can we deduce from this graph which method is better? (Thinking)