# Third Order ODE with unit step input

11 views (last 30 days)
Parveen Ayoubi on 27 Nov 2021
Commented: Walter Roberson on 28 Nov 2021
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.
Parveen Ayoubi on 28 Nov 2021
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

Walter Roberson on 27 Nov 2021
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 CommentsShowHide 1 older comment
Walter Roberson on 28 Nov 2021
However, if you take the laplace transform approach, then those are typically only valid for non-negative time.

R2021b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!