Question

The initial conditions are y(t = 0) = 2 and dy(t = 0)/dt = 0. The file VdP with the resulting right hand side
has been coded up for the given equation. Solve the equation for t = [0 : 0.01 : 32] using ode45. Is this the right code to solve for y(t)?
y0 = [2; 0];
[t,y1]=ode45(@VdP,[0:0.01:32],y0);
A4 = y1(:,1);

Answer 1

Matlab

The initial conditions are $y(t = 0) = 2$ and $dy(t = 0)/dt = 0$. The file VdP with the resulting right hand side has been coded up for the given equation. Solve the equation for $t = [0:0.01:32]$ using ode45. Is this the right code to solve for $y(t)$?

y0 = [2; 0];
[t,y1]=ode45(@VdP,[0:0.01:32],y0);
A4 = y1(:,1);

Yes, this code is right if code in the VdP-file is correct.

Here is an example of these two files for solving simple equation (with 0 initial cond.):

$$\frac{d^2 u}{dt^2} = 1$$

VdP:

function du=VdP(t,u)
du=zeros(2,1);
du(1)=u(2);
du(2)=1;

Solution:

[t,h]=ode45(@VdP,[0 300],[0,0]);
Solution = h(:,1)