almost-autnomous differential equation

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Bob 2016-7-23

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评论： Star Strider 2016-7-24

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.

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Star Strider 2016-7-23

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编辑：Star Strider 2016-7-23

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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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Bob 2016-7-24

Does this provide me with a value of c though.

Star Strider 2016-7-24

在 MATLAB Online 中打开

Not directly, but since I don’t understand what behaviour the particular value of ‘c’ is supposed to do, it should give you a way of determining the behaviour you’re looking for.

You can also plot the derivative (Jacobian) with meshc or contour if that would help:

t = linspace(0, 5, 50);
c = linspace(-3, 3, 50);
[T,C] = meshgrid(t,c);
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));
dfc_fcn = gradient(fc_fcn(T, C));
figure(1)
contour(T, C, dfc_fcn, 50)
grid on
xlabel('t')
ylabel('c')