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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