Error while outputting graph

Hello! Im currently running into a problem with getting my code to output my graph after updating to the most recent version of Matlabs and im unsure as to how to fix this as copilot seemingly is also not very helpful and im unsure how calling for subs works. Any help + explanation is greatly appreciated, thank you in advance!

Current Code:

syms y(t)

dy = diff(y)

dy2 = diff(dy)

ode = dy2 - dy == 2*t^2 - t - 5

ysol(t) = dsolve(ode)

c1 = 0;

c2 = 0;

ysol1(t) = subs(ysol(t) + c1*exp(t) + c2*t*exp(t))

c1 = -1;

c2 = -1;

ysol2(t) = subs(ysol + c1*exp(t) + c2*t*exp(t))

c1 = 3;

c2 = 3;

ysol3(t) = subs(ysol + c1*exp(t) + c2*t*exp(t))

figure

hold on

fplot(@(t) (ysol1(t)), [-2 2], '-', 'LineWidth', 1)

fplot(@(t) (ysol2(t)), [-2 2], ':', 'LineWidth', 1)

fplot(@(t) (ysol3(t)), [-2 2], '--', 'LineWidth', 1)

title('Problem One')

xlabel('t')

ylabel('Solutions')

grid on

ylim([-15 25])

legend('c1 = c2 =1 0','c1 = c2 = -1', 'c1 = c2 = 3')

Error:

Answer 1

[John D'Errico]

Not very clear exactly what you are trying to do. I'm pretty sure you want to substitute in different values of the undetermined constants, then plot each corresponding curve, but I'm often wrong.

syms y(t)
dy = diff(y);
dy2 = diff(dy);
ode = dy2 - dy == 2*t^2 - t - 5

ode(t) = 

ysol(t) = dsolve(ode)

ysol(t) = 

Now we can use subs.

ysol1 = subs(ysol,{'C1','C2'},[0,0])

ysol1(t) = 

Does that make sense? I told it to replace C1 and C2, with 0 and 0 respectively. Now do the same for the other choices of C1 and C2.

ysol2 = subs(ysol,{'C1','C2'},[-1,-1])

ysol2(t) = 

ysol3 = subs(ysol,{'C1','C2'},[3,3])

ysol3(t) = 

fplot(ysol1,'r')
hold on
fplot(ysol2,'g')
fplot(ysol3,'b')
legend('[0,0]','[-1,-1]','[3,3]')

Answer 2

[Star Strider]

You were not calling subs correctly. I added the necessary additional arguments, and it now seems to work.

Try this ---

syms y(t) C1 C2
dy = diff(y)

dy(t) = 

dy2 = diff(dy)

dy2(t) = 

ode = dy2 - dy == 2*t^2 - t - 5

ode(t) = 

ysol(t) = dsolve(ode)

ysol(t) = 

c1 = 0;
c2 = 0;
ysol1(t) = subs(ysol(t) + c1*exp(t) + c2*t*exp(t),  {C1,C2}, {0, 0})

ysol1(t) = 

c1 = -1;
c2 = -1;
ysol2(t) = subs(ysol + c1*exp(t) + c2*t*exp(t), {C1,C2}, {-1,-1})

ysol2(t) = 

c1 = 3;
c2 = 3;
ysol3(t) = subs(ysol + c1*exp(t) + c2*t*exp(t), {C1,C2}, {3,3})

ysol3(t) = 

figure
hold on
fplot(@(t) (ysol1(t)), [-2 2], '-', 'LineWidth', 1)
fplot(@(t) (ysol2(t)), [-2 2], ':', 'LineWidth', 1)
fplot(@(t) (ysol3(t)), [-2 2], '--', 'LineWidth', 1)
title('Problem One')
xlabel('t')
ylabel('Solutions')
grid on
ylim([-15 25])
legend('c1 = c2 =1 0','c1 = c2 = -1', 'c1 = c2 = 3')

.

Answer 3

[Torsten]

E.g.

syms y(t)
dy = diff(y);
dy2 = diff(dy);
ode = dy2 - dy == 2*t^2 - t - 5;
ysol(t) = dsolve(ode)

ysol(t) = 

s = symvar(ysol)  % Now you know that C1 = s(2) and C2 = s(3)

s = 

ysol1(t) = subs(ysol,[s(2),s(3)],[0 0]) % And here you substitute C1 = 0 and C2 = 0

ysol1(t) = 

fplot(@(t) (ysol1(t)), [-2 2], '-', 'LineWidth', 1) % And here you plot the corresponding graph