Variation of parameters. unsure why my solution differs from professor's

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SUMMARY

The discussion centers on solving the differential equation 2y'' - 3y' + y = ((t^2) + 1)e^t. The user's particular solution is (e^t) ((2/3)(t^3) + 6t - 4), while the professor's solution is ((1/3)(t^3)(e^t)) - 2(t^3)(e^t) + 9(te^t). A key insight provided by the user, ehild, indicates that the coefficient of y'' being non-unity necessitates dividing the right-hand side by 2 when calculating the functions u' and v'. This adjustment is crucial for arriving at the correct particular solution.

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Homework Statement


what is general solution of 2y'' - 3y' + y = ((t^2) + 1)e^t

Homework Equations



my particular solution is: (e^t) ((2/3)(t^3) + 6t -4))
prof particular solution is: ((1/3)(t^3)(e^t)) - 2(t^3)(e^t) + 9(te^t)

The Attempt at a Solution


here is how i solved , i hope this is ok to post link tp pdf..its faster than typing this all out..
https://docs.google.com/open?id=0BwJqUg33PgREQTFaejBCUkRRWlNfRW9maU1oaDNYQQ

thanks for any help/advice/hints
 
Last edited by a moderator:
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You can check if your solution is correct by substituting back into the original equation. ehild
 
As the coefficient of y'' is not 1 in the original equation, you have to divide the right-hand side by 2 when calculating u' and v'.

I think you did not copy your professor's solution correctly. It has to be
(1/3 t3-2t2+9t)et.

ehild
 
Last edited:

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