Given a function handle f an initial condition y0 and a final time tf, solve numerically the differential equation
dy/dt = f(y)
for the function y(t) between t=0 and t=tf. Give as a result res=y(tf).
Example:
f = @(x) -x; tf= 1; y0= 1;
=> y(tf) = 1/e = 0.367879441171442
Remarks: aim at a relative precision of around 1e-6. The function is analytic in the interval [0,1].
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last test needs a fix
@Nikolaos Nikolaou: the reference solution provided by the author passes the test suite, as do many community solutions. Can you clarify your comment?
As Nikolaos mentioned, the last test needs a fix. The value of 'a' is not transferred into the function handle 'f', and 'a' is just a char (not a value) in 'f'.
The last test case is not broken. When you define an anonymous function that includes variables, those variables are stored along with the function and remain available to it. This is true also when handles to the anonymous function are passed around, or when the variables in question are cleared from memory. See https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html#f4-71621 .
I am completely lost while solving it, please someone guide me