Dynamic Systems - DE with Unit Impulse

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SUMMARY

The discussion centers on solving a differential equation (DE) involving a unit impulse function, specifically represented by the Dirac delta function, δ(t). The equation is given as y(t) = e^{a(t-t_{0})} y(t_{0}) + ∫ e^{a(t-τ)} b ẋ(τ) dτ, with the solution requiring simplification under the condition c >> |a|. The final simplified form of the solution is y = (1/2) b c³ t² e^{(a-c)t} = b e^{at} δ(t). The participants explore the implications of removing variable 'a' from the solution and its effect on satisfying the DE.

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  • Understanding of differential equations, particularly linear first-order DEs.
  • Familiarity with the Dirac delta function and its properties.
  • Knowledge of integral calculus, especially evaluating integrals involving exponential functions.
  • Experience with MATLAB for numerical verification of solutions.
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Homework Statement


y(t) = e^{a(t-t_{0})} y(t_{0}) + \int_{t_{0}}^{t} e^{a(t-\tau)}b \dot{u} (\tau) d\tau
u(t) = \delta(t) = \frac{1}{2} c^{3} t^{2} e^{-ct}
where c >> |a|, t_0 = 0, and y(0) = 0
Find y(t) and represent the unit impulse, delta, in the solution. The remaining
terms should not contain c.

Homework Equations

The Attempt at a Solution


Evaluating the integral, I got
y = \frac{1}{2}bc^{3} \left( e^{-ct} \left(\frac{(ct^{2}-2t)(a+c)^{2}+(2ct-2)(a+c)+2c}{(a+c)^{3}} \right) + e^{at} \frac{2a}{(a+c)^{3}} \right)
However, I'm stuck at trying to substitute in delta to get a function that does not have the variable c in it. I'm assuming that since c >> |a|, then the terms with (a+c) will simplify to just c
y = \frac{1}{2}b \left( e^{-ct} \left((ct^{2}-2t)(c)^{2}+(2ct-2)(c)+2c \right) + e^{at} {2a}\right)
When I simplify it further, I am still unable to get it in a form without c. I'm thinking that I made a mistake somewhere in calculating the integral, but Matlab is giving me the similar result so I"m not sure where the problem is.
 
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I get something different when I evaluate the integral, but nevermind that. Try simplifying using c >> |a| before you evaluate the integral.
 
Doing that, I'm getting
y = \frac{1}{2} b c^{3} t^{2} e^{(a-c)t} = be^{at}\delta(t)
which is much simpler.

I just realized that I forgot to post the DE at the beginning, which is
\dot{y} = ay + b \dot{u}
When I try taking the derivative of y, I'm getting
\dot{y} = abe^{at}\delta(t) + be^{at}\dot{\delta}(t) = ay + be^{at} \dot{u}(t)
I'm assuming that since \dot{u}(t) has the term e^{-ct}, then the e^{at} will disappear, which would satisfy the DE.
Why is it that if I were to simplify y at the beginning to
y = \frac{1}{2} b c^{3} t^{2} e^{(-c)t} = b\delta(t)
and then taking the derivative,
\dot{y} = b\dot{u}(t)
Why is it that now it does not satisfy the DE, since ay is now missing? I'm curious as to why removing the variable a when it's being added/subtracted from c results in different forms depending on the step at which the simplification is done. Are all of the forms roughly equivalent to each other or something?

Thank you for the help.
 
Graphite said:
Why is it that now it does not satisfy the DE, since ay is now missing?
It does satisfy the DE, approximately, assuming ##c \gg |a|##. You can try plotting:
$$
\dot{y} = \frac{d}{dt}\left[be^{at} \delta(t)\right]\\
\dot{y} = \frac{d}{dt}\left[b \delta(t)\right]
$$
for some values of ##a,b,c##. Their graphs should be almost identical.

You have something "missing", because you've removed more information from the solution, comparatively - that has to be reflected somewhere.
 

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