Is it possible to get a time-dependent function as an output of an ODE solver?

My code is below, it plots as I expect it to. I would like to know if it is possible to get an output where the ODE solver gives the solution as a function in time rather than 2 column vectors [T,Y]. I wanted to use the output to further analyze something, but need a function in time rather than the column vectors.

clear all

clc

I0= 500; % maximum flow

Tc=60/72; % heart period

Ts=(2/5*Tc); % time in systole

P_ss=80; % diastolic pressure

R= 1;

C=1;

R1=0.05;

L=0.005;

I=@(t)I0*sin((pi*t)/Ts).^2.*(t<=Ts); %input current flow

Idot = @(t)I0*2*sin(pi*t/Ts).*cos(pi*t/Ts)*pi/Ts.*(t<=Ts);

Idotdot = @(t) I0*2*(cos(pi*t/Ts).^2 - sin(pi*t/Ts).^2)*(pi/Ts)^2.*(t<=Ts);

fun = @(t,y)[y(2);(Idotdot(t)*(R*L*C*R1)+Idot(t)*(L*(R+R1))+I(t)*(R*R1) - (y(2)*(C*R*R1+L)+y(1)*R1))/(L*C*R)];

y0 = [80 ;0];

tspan = [0 60/72];

[T,Y] = ode45(fun,tspan,y0);

plot(T,Y(:,1),'g')

Answer 1

[Torsten]

clear all
clc   
I0= 500; % maximum flow
Tc=60/72;   % heart period
Ts=(2/5*Tc);   % time in systole
P_ss=80; % diastolic pressure
R= 1;
C=1;
R1=0.05;    
L=0.005;
I=@(t)I0*sin((pi*t)/Ts).^2.*(t<=Ts); %input current flow
Idot = @(t)I0*2*sin(pi*t/Ts).*cos(pi*t/Ts)*pi/Ts.*(t<=Ts);
Idotdot = @(t) I0*2*(cos(pi*t/Ts).^2 - sin(pi*t/Ts).^2)*(pi/Ts)^2.*(t<=Ts);
fun = @(t,y)[y(2);(Idotdot(t)*(R*L*C*R1)+Idot(t)*(L*(R+R1))+I(t)*(R*R1) - (y(2)*(C*R*R1+L)+y(1)*R1))/(L*C*R)];
y0 = [80 ;0];
tspan = [0 60/72];
[T,Y] = ode45(fun,tspan,y0);
plot(T,Y(:,1),'g')

function_of_time = @(t)interp1(T,Y(:,1),t);
function_of_time(30/72)

ans = 119.2901

Answer 2

[Steven Lord]

Call ode45 with one output argument. Pass that output from ode45 and an array of times into the deval function to evaluate the solution at those times.