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# reviewsolutions - 2 = 1 4-1 3 1 2 C so C = 1 12 Plugging...

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2.1.15 Solve the ODE ty 0 + 2 y = t 2 - t + 1 , y (1) = 1 2 , t > 0 . We start by dividing by t to transform the equation into one where we can use the method of integrating factors: y 0 + 2 t y = t - 1 + 1 t . (1) Thus μ ( t ) = exp Z 2 dt t = e 2 ln | t | = t 2 . Multiplying both sides of (1) by μ , we find t 2 y 0 + 2 ty | {z } d dt ( t 2 y ) = t 3 - t 2 + t. Thus t 2 y = t 4 4 - t 3 3 + t 2 2 + C. (2) Plugging in our initial condition y (1) = 1

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Unformatted text preview: / 2 = 1 / 4-1 / 3+1 / 2+ C , so C = 1 / 12. Plugging our value for C into (2) and dividing both sides by t 2 , we conclude y = t 2 4-t 3 + 1 2 + 1 12 t 2 . 3.3.16 By Abel’s Theorem the Wronskian of the Above equation is …...
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reviewsolutions - 2 = 1 4-1 3 1 2 C so C = 1 12 Plugging...

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