# [SOLVED] really complex looking differential equation!!

• Feb 4th 2009, 08:12 AM
crafty
[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
• Feb 4th 2009, 09:49 AM
Jester
Quote:

Originally Posted by crafty
(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?
• Feb 4th 2009, 01:28 PM
crafty
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?
• Feb 4th 2009, 03:36 PM
crafty
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?
• Feb 4th 2009, 03:48 PM
Jester
Quote:

Originally Posted by crafty
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?

$\displaystyle \frac{d}{dt} \left( \frac{y}{t} \right) = t \sin t$

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

$\displaystyle y = t \sin t - t^2 \cos t + c \,t$
• Feb 4th 2009, 04:04 PM
crafty
i made the same mistake twice ... our lecturer hasnt gone through much about the constant and how its affected ..
• Feb 4th 2009, 04:09 PM
crafty
so if i'm given an initial value y(pi) =0 then is the value of c = -pi
• Feb 4th 2009, 05:46 PM
Jhevon
Quote:

Originally Posted by crafty
so if i'm given an initial value y(pi) =0 then is the value of c = -pi

yes