(t^2-1).y''+ty'+3y=t y(0)=1/2 y'(0)=3
In view of the initial conditions, the series starts as y(t) = 1/2 + 3t + a_{2}t^{2} + a_{3}t^{3} + a_{4}t^{4 }+ a_{5}t^{5} + …
Differentiation yields y'(t) = 3 + 2a_{2}t + 3a_{3}t^{2} + 4a_{4}t^{3 }+ 5a_{5}t^{4} + … ;
y''(t) = 2a_{2} + 6a_{3}t + 12a_{4}t^{2 }+ 20a_{5}t^{3} + … .
Substitute the obtained expressions into the equation:
(t^{2} – 1) (2a_{2} + 6a_{3}t + 12a_{4}t^{2 }+ 20a_{5}t^{3} + …) + t(3 + 2a_{2}t + 3a_{3}t^{2} + 4a_{4}t^{3 }+ 5a_{5}t^{4} + …) + 3(1/2 + 3t + a_{2}t^{2} + a_{3}t^{3} + a_{4}t^{4 }+ a_{5}t^{5} + …) = t and compare the coefficients at the powers of t
t^{0}: –2a_{2 }+ 3/2 = 0 → a_{2 }= 3/4
t^{1}: – 6a_{3 }+ 3 + 9 = 1 → a_{3 }= 11/6
t^{2}: 2a_{2}– 12a_{4 }+ 2a_{2} + 3a_{2} = 0 → a_{4 }= 7a_{2 }/12 = 7/16
t^{3}: 6a_{3}– 20a_{5 }+ 3a_{3} + 3a_{3} = 0 → a_{5 }= 12a_{3 }/20 = 11/10
Finally, y(t) = 1/2 + 3t + (3/4)t^{2} + (11/6)t^{3} + (7/16)t^{4 }+ (11/10)t^{5} + …
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