# Thread: Particular and transient solutions for 2nd order differential equations.

1. ## Particular and transient solutions for 2nd order differential equations.

Ok, so I am doing some basic circuit analysis and it's been awhile since I have done any diff. eq's so I need a little help here.

I have an RLC circuit with a 2.5 Ohm resistor, a 1 Henry Inductor and a 1 farad Capacitor all in series. With an input of 24*u(t) where u(t) is the unit step function.

So I derived the 2nd order differential equation of.

y'' + 2.5y' + y = 24

Now the transient solutions will be found with the characteristic equation:

r^2 + 2.5r +1 = 0

giving me roots of -.5 and -2 and a transient function:

y(t) = c1*e^-.5t + c2*e^-2t

Now I can't remember how to find the Particular solution, is it:

Y(t) = At*e^-t

??
then find Y'(t) and Y''(t) don't really remember where to go from there.

Am I moving in the right direction?

2. I would just go with a particular solution of y = 24. It works, doesn't it? The derivatives vanish, and what's left satisfies the DE.

3. Ok. I see, since it is a constant you know that the solution is just 24.

But if the input was 24sint. You would have to have a particular solution of

Asint + Bcost

and if it was 24e^t. you would have to have

Ate^t

Is that right? It's starting to come back to me a little bit.

4. It depends on the transients. If the homogeneous solution has terms that look like the RHS, then you have to multiply by t (you're essentially using variation of parameters) in order to avoid having the operator represented on the LHS "kill" your particular solution. You have good guesses there for particular solutions.

5. Ok. thats right.

Thanks.

6. You're welcome. Have a good one!

7. Ok so I just want to make sure that I have a complete grasp on this.

I solved and got a transient solution of y(t)=32e^-.5t - 8e^-2t
and the particular solution of Y(t)=24
so for the total output I thought you added the two together which would be 32e^-.5 - 8e^-2t + 24
but this function starts at 48 and exponentially decays to 24, but thinking it through logically shouldn't it be the opposite start at 0 and exponentially grow to 24.

8. It looks like you applied the initial conditions to the homogeneous solution. I think you have to apply them to the general solution. What are the initial conditions, anyway?

9. The only thing that is given is that the input is 24*u(t).

so it is zero when t is less than 0 and 24 when t>=0.

10. Oh. Then you don't know enough to determine the c1 and c2 of the OP. You have to leave those arbitrary constants in there.

11. That was my initial thought but he made the problem out as if we could solve for a solution and plot it as a function of time.

12. Hmm. I don't think I agree that it can be done. Incidentally, I think I may be missing something in the solution. Here's WolframAlpha's solution (Mathematica gives the same).

13. hmm....that confuses me further.

14. Yeah, me, too. I'm wondering if the approach we've taken has to be modified: solve two DE's, one for $\displaystyle t<0$ and one for $\displaystyle t\ge 0.$ That would explain the extra terms. Since the extra terms are all multiplied by the step function, they don't kick in until $\displaystyle t\ge 0.$ Maybe Danny can explain it better.

15. ## update.

Ok. So he left out that he wanted us to assume that the circuit had been at rest for a long period of time. This would mean that the initial conditions would be zero for y(0) and y'(0) (the initial voltage and current across the capacitor and inductor respectively). Also I think that we are still correct with using 24 as the particular solution, since my problem is only worried about t>0. Dealing with t<0 makes for a much messier situation probably looking like the Mathematica solution. Side question about that, would that have to be solved using Laplace Transformations? I haven't had any experience with these yet and they look pretty messy.

Anyway, my final answer comes to:

-32e^-.5t + 8e^-2t +24

which when plotted looks and acts as it should, so I think I got it right.

Thanks for all the help on this one.

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