MAT197 – Calculus B
Winter 2014
Solving Differential Equations using Power Series
We can use Taylor series to solve differential equations.
Example. Take the following LRC−series circuit. We have
• E(t) = impressed voltage
inductor
• i(t) = current
• q(t) = charge of the capacitor:
dq
dt
L
=i
E
• voltage drop across resistor1 : Ri = R dq
dt
R
resistor
C
2
di
• voltage drop across inductor2 : L dt
= L ddt2q
3
• voltage drop across capacitor :
capacitor
1
Cq
Kirchhoff ’s Second Law: Sum of the voltage drops over the components of the circuit is equal to the
impressed voltage.
L
d2 q
dq
q
+R +
= E(t),
dt2
dt
C
where E(t) is the impressed voltage.
Example. A 12-volt battery is connected to a single loop LRC-circuit with L =
1
4
Henry, R = 2 ohms, C =
Farads. Assuming that q(0) = i(0) = 0, determine the charge q(t).
Solution. The DE we want to solve is


 14 q 00 (t) + 2q 0 (t) + 4q = 12


q(0) = 0



q 0 (0) = i(0) = 0
The idea is to assume that the solution can be written as a power series:
q(t) =
∞
X
ck tk .
k=0
7A
resistor is a passive two-terminal electrical component that implements electrical resistance as a circuit element.
inductor is a passive two-terminal electrical component which resists changes in electric current passing through it.
9 A capacitor is a passive two-terminal electrical component used to store energy electrostatically in an electric field.
8 An
1
1
4
MAT197 – Winter 2014
Then we can find the derivatives
q 0 (t) =
q 00 (t) =
∞
X
k=1
∞
X
kck tk−1 =
∞
X
ck+1 (k + 1)tk
k=0
k(k − 1)ck tk−2 =
k=2
∞
X
ck+2 (k + 1)(k + 2)tk
k=0
We now use these series and sub them into the DE:
∞
∞
∞
X
X
1X
ck+2 (k + 1)(k + 2)tk + 2
ck+1 (k + 1)tk + 4
ck tk = 12
4
k=0
k=0
k=0
∞ X
(k + 1)(k + 2)
ck+2 + 2(k + 1)ck+1 + 4ck tk = 12
4
k=0
Just like with polynomials, the coefficients of the same power of t have to match, so we have
c2
+ 2c1 + 4c0 = 12
2
(t0 )
(tk )
(k + 1)(k + 2)
ck+2 + 2(k + 1)ck+1 + 4ck = 0
4
for k = 1, 2, . . .
The second equation tells us that
ck+2 = −
8
16
ck+1 −
ck
(k + 2)
(k + 1)(k + 2)
for k = 1, 2, . . . .
The first equation states that
c2 = 24 − 4c1 − 8c0 .
This defines the power series of the solution recursively.
Q.
How do we find c0 and c1 ?
We use the initial conditions:
0 = q(0) = c0
⇒
c0 = 0
0 = q 0 (0) = c1
⇒
c1 = 0
c3 = −64
c4 = 128 − 32 = 96
This means that
c2 = 24
,
,
So the solution is
q(t) = 24 t2 − 64 t3 + 96 t4 + · · ·
2
,
···
MAT197 – Winter 2014
Notes.
1.
This may seem like a “bad” way to write the solution, but we can identify some behaviour of the solution
like this, like for example:
q 00 (0)
i0 (0)
=
= 24
2!
2!
000
00
q (0)
i (0)
=
= −64
3!
3!
⇒
i0 (0) = 48
⇒
i00 (0) = −384
So we know that at the beginning, the current is increasing, but it’s “decelerating” (its rate of increase is
going down).
2.
We can also differentiate q(t) term-by-term to obtain an expression for i(t):
i(t) = q 0 (t) = 48t − 192t2 + 384t3 + · · ·
3