Saturday, August 17, 2013

Solutions to Applications of Higher Order Differential Equations (2) - Part B

2) For the circuit shown, consider the special case of NO capacitor. Then the differential equation of interest becomes:

L (dI/dt) + RI = E

It’s instructive to first solve this form where E is presumed constant. Begin by re-arranging the DE to get:

L dI = (E – RI) dt so that:  ò LdI/ (E – RI) = ò  dt

Let u = (E – RI) ,    Then: du = - RdI and dI = -du/ R

So:   - L/R  ò du/u = ò  dt
Now, u = E – RI and this has a (+ve) value (e.g. uo = E) at t = 0 and we assume it remains positive so çu ç = u and:
(-L/R) ln u  = t + C
Or: -(L/R) ln (E – RI)  = t + C
The constant of integration C can be found knowing that I = 0 at t = 0 hence:
C = - (L/R)  ln E

After suitable substitutions and some algebra:

L/R [ln E – ln (E – RI)] = t
ln [E/ E – RI]  = Rt/ L
Taking natural logs:
E/ (E – RI)  = exp (Rt/L)
Which can also be written:  (E – RI)/ E = exp (- Rt/L)

Solving for I:   I = E/R [1 - exp (-Rt/L)]
Then the current (for t > 0) always increases toward the steady state value of E/R. Now, what about the solution if we let E = Eo sin wt? (Oscillatory voltage). 
It should be fairly clear from past articles –blog posts that the solution must have the form: A sin wt + B cos wt
Using the integrating factor, the solution we obtain (note this is not a higher order DE!) is:
I exp (at)  =  ò exp (at)  E dt  + C
Then one will obtain:
I =   Eo { RL sin wt -  wL 2 cos wt}/ (R2 +  w2 L 2   )  +  k exp (-Rt/L)
This can also be done using the Laplace transform technique, after re-arranging the equation to get:

dI/dt + (R/L) I =  Eo sin wt
Then write the transformed eqn. as, using  a = R/L:
s I(s) + a I(s) =   Eo  w/ (s2 +  w2 )
I(s) [ s + a] =   Eo  w/ (s2 +  w2 )
I(s) =  Eo  w  / (s + iw) (s - iw)  (s + a)
For which a partial fractions decomposition can be used, followed by the inverse Laplace transform.







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