Help solving a second order differential equation

clear;clc
syms y(t)



fun = 0.001*diff(y,t,2)+(1050)*diff(y,t)+(1/0.0047)*y == 0;

cond1 = y(0) == 0;
cond2 = diff(y) == 0;

conds = [cond1 cond2];

ySol(t) = dsolve(fun,conds);
%ySol(t) = dsolve(fun);
ySol = simplify(ySol);
disp(ySol(t))

When I run the code I get the following error: "Unable to reduce to square system because the number of equations differs from the number of indeterminates."

Thank you.

Answer 1

[Star Strider]

If you use the numeric initial conditions, you get the trivial solution only, that being 0.

If you want to see the full expression (you can substitute in for the initial conditions later), this woirks:

syms y(t) y0 Dy0 
Dy = diff(y,t);
D2y = diff(y,t,2);
fun = 0.001*D2y == -((1050)*Dy+(1/0.0047)*y);
cond1 = y(0) == y0;
cond2 = Dy(0) == Dy0;
conds = [cond1 cond2];
ySol(t) = dsolve(fun,conds);
%ySol(t) = dsolve(fun);
ySol = simplify(ySol, 'Steps',20)
disp(ySol(t))

producing:

(608855155^(1/2)*exp(t*((1000*608855155^(1/2))/47 - 525000))*(47*Dy0 + 24675000*y0 + 1000*608855155^(1/2)*y0))/1217710310000 - exp(-t*((1000*608855155^(1/2))/47 + 525000))*((608855155^(1/2)*Dy0)/25908730000 - y0/2 + (105*608855155^(1/2)*y0)/5181746)

[Reymi Chacon]

How do I define the initial conditions after the equation has been solved? I tried

y0=0;
dy0=0;

But it doesnt work. Thanks fot the reply btw

[Star Strider]

My pleasure.

Use the subs function:

ySol = subs(ySol, {y0, Dy0}, {0, 0})

The result is still 0 if you do that.