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Math Help - [SOLVED] really complex looking differential equation!!

  1. #1
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    [SOLVED] really complex looking differential equation!!

    (t)*(dy/dt) - y = (t^3)(sin t) and then use initial value y(pi) = 0 to find the constant of integration the answer i got for C was ((pi^2) - 2)/pi can someone verify this for me?? and also the answer i got was y = (1/t)*[ (t^2)(cos t) + (2t)(sin t) - (2cos t) + C] i used the integrating factor method
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    Quote Originally Posted by crafty View Post
    (t)*(dy/dt) - y = (t^3)(sin t) and then use initial value y(pi) = 0 to find the constant of integration the answer i got for C was ((pi^2) - 2)/pi can someone verify this for me?? and also the answer i got was y = (1/t)*[ (t^2)(cos t) + (2t)(sin t) - (2cos t) + C] i used the integrating factor method
    I got a different answer. Can you show some steps?
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    well after rearranging t and y, I got my answer in the form: dy/dt = (1/t)y +(t^2)(sin t) dt then i got my integrating factor as t and after putting substituting in the equation i got the answer i posted where did i go wrong? can you show me my mistake?
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    after trying it again i think i caught my mistake: i had an extra negative sign while finding out the integrating factor: after repeating it i got the integration factor to be (1/t) and my final answer is: y = (t^2)(cos t) + (t)(sin t) + C this has to be correct? am i right?
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    Quote Originally Posted by crafty View Post
    after trying it again i think i caught my mistake: i had an extra negative sign while finding out the integrating factor: after repeating it i got the integration factor to be (1/t) and my final answer is: y = (t^2)(cos t) + (t)(sin t) + C this has to be correct? am i right?
     \frac{d}{dt} \left( \frac{y}{t} \right) = t \sin t

     \frac{y}{t} = \sin t - t \cos t + c

     y = t \sin t - t^2 \cos t + c \,t
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    i made the same mistake twice ... our lecturer hasnt gone through much about the constant and how its affected ..
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  7. #7
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    so if i'm given an initial value y(pi) =0 then is the value of c = -pi
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by crafty View Post
    so if i'm given an initial value y(pi) =0 then is the value of c = -pi
    yes
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