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Differential Equations Variation of Params Problems

  • Thread starter lubuntu
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  • #1
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Homework Statement



[tex]t^{2}[/tex] * y'' - 2y = [tex]3t^{2}[/tex] - 1

[tex]y_{1}[/tex] = t[tex]^{2}[/tex]

[tex]y_{2}[/tex] = 1/t


Homework Equations



Variation of Params forumla
wronskian det



The Attempt at a Solution



W = t^2 * -t^-2 -[1/t * 2t] = -3

Y = -t^2 * Integral[ (-1/3)(1/t)(3t^2 -1)] + 1/t * Integral[(-1/3)(t^2)(3t^2 -1)]

I get an answer that is not correct so I guess I set this part up wrong somehow anyone see the error?

My answer: Y = 3/10*t^4 - t^2 * ln|t|/3

Correct Answer: Y = t^2*ln|t| + 1/2
 
Last edited:

Answers and Replies

  • #2
459
1
So I figured I ended up getting the correct answer but going the long way and solving the system using

u*t^2 + v * 1/t = 3t^2 -1

but I still have no idea why the integral gave me the wrong answer apparently I missed something somewhere, if you feel like going through please let me know what I messed up on.
 

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