How is this an exponential curve?

Hello,

I would like your help on the following code.

How does this trace an exponential curve? N=8000; % number of steps to take T=8; % maximum time h=T/N; % time step t=(0:h:T); % t is the vector [0 1h 2h 3h ... Nh] y=zeros(size(t)); % prepare place to store locations

y(1)=3;             % initial height
for i=1:N           % start taking steps
  y(i+1)=y(i)-y(i)*h;
end;
plot(t,y), hold on  % plot more permanently

y(1)=-2;            % initial height
for i=1:N           % start taking steps
  y(i+1)=y(i)-y(i)*h;
end;
plot(t,y);          % plot more permanently

Please help.

Thanks

Answer 1

[Shoaibur Rahman]

Convert your difference equation into differential equation:

y(i+1) = y(i)-y(i)*h;
y(i+1) - y(i) = -y(i)*h;
(y(i+1) - y(i))/h = -y(i);
dy/dt = -y

Time step h is replaced by dt to make the equation more understandable. Now, solve this differential equation using variable separation method:

dy/y = -dt
ln|y| = -t + C
y = exp(-t+C)
y = A*exp(-t)

At t = 0, A = y(0) = 3, in your case, y(0) is equivalent to y(1) in Matlab

y = 3*exp(-t)

This means that y is exponential. However, when solving using difference equation, make sure that h or the time is small enough to keep the solution stable. For this particular equation, h<1 . Use a larger value of h in your code just to see the effect.

Answer 2

[Torsten]

The above code is an implementation of the explicit Euler method to solve the differential equations

y'=-y, y(0)=3

y'=-y, y(0)=-2

The above differential equations have solutions

y(t)=3*exp(-t)

y(t)=-2*exp(-t)

So yes, the program generates approximations to exponentially decaying curves.

Best wishes

Torsten.