MATLAB: Almost-autnomous differential equation

differential equation

y' = y^2 − t. This differential equation has “stationary” solutions, but unlike with an autonomous equation, those stationary solutions are not horizontal. Vary the initial condition y(0) = c for a bit to try to get a sense of what the solutions look like. (Picking values between −3 and +3 should be good enough.)
Part A: There’s a value P such that, if y(0) < P, then the solution to the initial value problem decreases, while if y(0) > P, the solution to the initial value problem increases. Figure out what P is to two decimal places.
Attempted code:
    syms y(t);
    for c = -3:1:3
        fc = dsolve('Dy = y^2 - t' , y(0) == c);
    end
Not sure how I get it to print out an answer for p and get it to be for p to two decimal places.

Best Answer

You need to solve it symbolically, but then you can use the matlabFunction function to create an anonymous function from it:
syms y(t) c
fc(t,c) = dsolve(diff(y) == y^2 - t, y(0) == c);
fc_fcn = matlabFunction(fc)
fc_fcn = @(t,c) -(airy(3,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(1,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0))./(airy(2,t)+((sqrt(3.0).*gamma(2.0./3.0).^2.*3.0+3.0.^(2.0./3.0).*c.*pi.*2.0).*airy(0,t))./(gamma(2.0./3.0).^2.*3.0-3.0.^(1.0./6.0).*c.*pi.*2.0));
You also need to decide on an appropriate range for ‘t’ if you haven’t already been given one. It might be easier to use a ribbon plot for this until you home in on the correct value for ‘c’.
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