Third Order ODE with unit step input

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Parveen Ayoubi
Parveen Ayoubi 2021-11-27
评论： Walter Roberson 2021-11-28
I have been trying to solve this differential equation for two days now. I do not know what to do with the right hand side of the ODE. The only way I have seen to solve it does not include the derivative of the input as well. Would really appreaciate some help atleast to know how to start it up.
y^''' (t)+6y^'' (t)+11y^'(t) +6y(t)=u^'' (t)+2u^' (t)+3u(t)
y’’(0) = 1 ; y’(0) = -1; y(0) = 1
where u=Unit step Us(t).
Ive tried to do it in simulink but the answers there havent been coming out right either.
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Parveen Ayoubi
Parveen Ayoubi 2021-11-28
Wouldnt you be missing the initial conditions of U when changing from time domain to frequency? I think that is the difference, Because if we have unit step for the input then u(0)=1 u'(0) is equal to one as well. If you can show me how i would add those conditions to the code you did it would be great. Or atleast a hint
Parveen Ayoubi
Parveen Ayoubi 2021-11-28
Nevermind im incorrect, which makes all my work wrong and im at another wall. Because if im assuming that t>0 like walter robinson had in his code it wouldnt be the correct answer

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Walter Roberson
Walter Roberson 2021-11-27
Ran in:
sympref('HeavisideAtOrigin', 0);
syms y(t)
assume(t > 0)
Dy = diff(y,t);
D2y = diff(y,t,2);
u(t) = heaviside(t); %// I saw online that this is unit step
Du = diff(u,t);
ode = diff(y,t,3) + 6*diff(y,t,2)+11*diff(y,t)+6*y == diff(u,t,2)+2*diff(u,t)+3*u;
cond1 = Dy(0) == -1;
cond2 = D2y(0) == 1;
cond3 = y(0) == 1;
conds = [cond1 cond2 cond3];
sol = dsolve(ode,conds)
sol = 
The assumption that t > 0 is not strictly valid, but if you use t >= 0 you get sign(t) and heaviside() in the equations, and you would have to break it into two cases. You could continue on with
assume(t == 0)
sol0 = dsolve(ode,conds)
sol0 = 
... which happens to give the same result.
  2 个评论
Walter Roberson
Walter Roberson 2021-11-28
Ran in:
sympref('HeavisideAtOrigin', 0);
syms y(t)
assume(t,'real');
Dy = diff(y,t);
D2y = diff(y,t,2);
u(t) = heaviside(t); %// I saw online that this is unit step
Du = diff(u,t);
ode = diff(y,t,3) + 6*diff(y,t,2)+11*diff(y,t)+6*y == diff(u,t,2)+2*diff(u,t)+3*u;
cond1 = Dy(0) == -1;
cond2 = D2y(0) == 1;
cond3 = y(0) == 1;
conds = [cond1 cond2 cond3];
sol = simplify(dsolve(ode,conds))
sol = 
sol0 = limit(sol, t, 0)
sol0 = 
1
syms tpos tneg
assume(tpos > 0)
assume(tneg < 0)
solpos = subs(simplify(subs(sol, t, tpos)),tpos,t)
solpos = 
solneg = subs(simplify(subs(sol, t, tneg)),tneg,t)
solneg = 
Walter Roberson
Walter Roberson 2021-11-28
However, if you take the laplace transform approach, then those are typically only valid for non-negative time.

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