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Use the series solution method to find the first six terms of the power series for solution to IVP

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(t^2-1).y''+ty'+3y=t      y(0)=1/2  y'(0)=3

Apr 9th, 2015

In view of the initial conditions, the series starts as y(t) = 1/2 + 3t + a2t2 + a3t3 + a4t4 + a5t5 + …

Differentiation yields y'(t) = 3 + 2a2t + 3a3t2 + 4a4t3 + 5a5t4 + … ;

y''(t) = 2a2 + 6a3t + 12a4t2 + 20a5t3 + … .

Substitute the obtained expressions into the equation:

(t2 – 1) (2a2 + 6a3t + 12a4t2 + 20a5t3 + …) + t(3 + 2a2t + 3a3t2 + 4a4t3 + 5a5t4 + …) + 3(1/2 + 3t + a2t2 + a3t3 + a4t4 + a5t5 + …) = t and compare the coefficients at the powers of t

t0: –2a2 + 3/2 = 0 → a2 = 3/4

t1: – 6a3 + 3 + 9 = 1 → a3 = 11/6

t2: 2a2– 12a4 + 2a2 + 3a2 = 0 → a4 = 7a2 /12 = 7/16

t3: 6a3– 20a5 + 3a3 + 3a3 = 0 → a5 = 12a3 /20 = 11/10

Finally, y(t) = 1/2 + 3t + (3/4)t2 + (11/6)t3 + (7/16)t4 + (11/10)t5 + …


Apr 9th, 2015

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