Why I am I getting a row

How should I fix it so that x1 can be a column

r = 1;

K = 2;

tspan = [0 10];

x0 = 0.5;

[t,x] = ode45(@(t,x) x*(1- x/K), tspan, y0);

f = @(t) K / (1 + (K / x0 - 1) * exp(-1*r*t));

x1 = f(t);

Answer 1

[Star Strider]

Thge easiest way is:

x1 = x1(:)

since this will force ‘x1’ to be a column vector regardless of its previous orientation.

r = 1;
K = 2;
tspan = [0 10];
x0 = 0.5;
[t,x] = ode45(@(t,x) x*(1- x/K), tspan, x0);
f = @(t) K / (1 + (K / x0 - 1) * exp(-1*r*t));
x1 = f(t);
x1 = x1(:)

x1 = 45×1

0.5000 0 0 0 0 0 0 0 0 0

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[Zheng]

Hi, I intend to input this column vector of t as well in order to not getting zeros. How should I fix it? Thank you!

[Star Strider]

I am not certain what you are asking.

In order to not produce any zeros, your ‘f(t)’ function should not produce them.

r = 1;
K = 2;
tspan = [0 10];
x0 = 0.5;
[t,x] = ode45(@(t,x) x*(1- x/K), tspan, x0);

figure
plot(t,x)
grid

●

f = @(t) K / (1 + (K / x0 - 1) * exp(-1*r*t));
x1 = f(t);
x1 = x1(:)

x1 = 45×1

0.5000 0 0 0 0 0 0 0 0 0

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<mw-icon class=""></mw-icon>

.

[Zheng]

For x1, I am using a different formula f = @(t) K / (1 + (K / x0 - 1) * exp(-1*r*t)). If you try to input index like f(1), f(2), it outputs a number perfectly well and it should not be 0. I think there is a problem of how this formula is defined.

[Star Strider]

‘I think there is a problem of how this formula is defined.’

I agree. However since your coding for it is all that I have to work with, this is the best that I can do with it. Check to be certain that ‘r’ and ‘K’ have the correct values.

r = 1;
K = 2;
tspan = [0 10];
x0 = 0.5;
[t,x] = ode45(@(t,x) x*(1- x/K), tspan, x0);

figure
plot(t, x, '.-')
grid

disp(t)

         0
    0.0670
    0.1340
    0.2010
    0.2679
    0.5179
    0.7679
    1.0179
    1.2679
    1.5179
    1.7679
    2.0179
    2.2679
    2.5179
    2.7679
    3.0179
    3.2679
    3.5179
    3.7679
    4.0179
    4.2679
    4.5179
    4.7679
    5.0179
    5.2679
    5.5179
    5.7679
    6.0179
    6.2679
    6.5179
    6.7679
    7.0179
    7.2679
    7.5179
    7.7679
    8.0179
    8.2679
    8.5179
    8.7679
    9.0179
    9.2679
    9.4510
    9.6340
    9.8170
   10.0000

f = @(t) K / (1 + (K / x0 - 1) * exp(-1*r*t));
x1 = f(t);
x1 = x1(:);
disp(x1)

    0.5000
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0
         0

clearvars

syms K r t x(t) x0 
DE = diff(x) == x*(1- x/K)

DE(t) = 

f = dsolve(DE, x(0) == x0)

f = 

f = simplify(f, 500)

f = 

fcn = matlabFunction(f)

fcn = function_handle with value:
    @(K,t,x0)(K.*x0.*exp(t))./(K-x0+x0.*exp(t))

t = (0:10).';
K = 2;
x0 = 0.5;
fv = fcn(K,t,x0);
disp(fv)

    0.5000
    0.9507
    1.4225
    1.7401
    1.8958
    1.9604
    1.9852
    1.9945
    1.9980
    1.9993
    1.9997

figure
plot(t, fv, '.-')
grid
xlabel('t')
ylabel('f(t)')

For what it is worth, the analytiic integration of your function does not closely resemble your anonymous function formulation of it.

.