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Initial conditions on ODE45 ?

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Robin
Robin 2011-7-30
评论： Pasindu Ranasinghe 2021-7-22
Im trying to solve this IVP: e^y +(t*e^y - sin(y))*(dy/dt)=0 with the initial condition y(2)=1.5.
I was just not sure how to do it with the initial condition with Y(2)=1.5, iknow how to do it if it were y(0)=1.5:
f= @(t,y) (exp(y)+(t.*exp(y)-sin(y))); % This is the function.
[t,y]=ode45(f, [0.5,4], 1.5); % trange is from 0.5 to 4
plot(t,y)
can someone please help me out?

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采纳的回答

Jan
Jan 2011-7-30
This uses the initial value y(0.5)=1.5 ( not y(0)=1.5):
[t, y] = ode45(f, [0.5, 4], 1.5);
So for y(2)=1.5:
[t, y] = ode45(f, [2, 4], 1.5);
Note: The initial value problem starts at the inital point.
[EDITED]: The call to ODE45 is equivalent, if the problem is formulated in backward direction - an "final value problem": tspan is still [ti, tf], but now ti > tf.
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显示 7更早的评论
Walter Roberson
Walter Roberson 2017-7-26
Liu Langtian comments to Jan Simon
right
Pasindu Ranasinghe
Pasindu Ranasinghe 2021-7-22
Example Code
Use ode45() to find the approximate values of the solution at t in the range of 1 to 3
function ydot = eqns(t,y)
ydot=(t-exp(-t))/(y+exp(y));
end
###################################
%%Code
[t1,y1]=ode45(@eqns,[1.5 1], 0.5);
hold on;
[t2,y2]=ode45(@eqns,[1.5 3], 0.5);
hold off
t=[t1;t2];
y=[y1;y2];
plot(t,y,'-o')

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更多回答（1 个）

Subha Fernando
Subha Fernando 2011-10-26
let say function is dy/dt = y (t-y).
If initial condition is given at y(1) = 0.5 not at y(0) then we define the RHS as
function output = funcRHS(t, y) output = y *(t-y); end
%then u can call
hold on ode45('funcRHS', [1, -1], 0.5) ode45('funcRHS', [1,5], 0.5)
%Here you can see and read the initial value at y(0) also

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