Help solving a second order differential equation
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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.
回答（1 个）
Star Strider
2018-12-6
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)
2 个评论
Reymi Chacon
2018-12-6
Star Strider
2018-12-6
My pleasure.
Use the subs function:
ySol = subs(ySol, {y0, Dy0}, {0, 0})
The result is still 0 if you do that.
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2018-12-6
2018-12-6
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