Please check my work (differential equation)

[darryw]

Homework Statement

ty' + 2y = sin t (no initial conditions given)

Homework Equations

The Attempt at a Solution

ty' + 2y = sin t

y' + (2/t)y = sin t / t

mu(x) = e^integ(2/t) = t^2

(t^2)y)' = integ t sin t

(t^2)y = tsin t - t cos t + c

y = (sin t - cos t)/ t + c/(t^2) (This is my solution)

thanks for any help

 

[tiny-tim]

Hi darryw!

(have an integral: ∫ and a mu: µ and try using the X² tag just above the Reply box )

Your equations are ok down to …

(t^2)y)' = integ t sin t

(t^2)y = tsin t - t cos t + c

y = (sin t - cos t)/ t + c/(t^2)

… the first line of course should be (t^2)y)' = t sin t (without the ∫) ,

but more seriously your integration by parts has come out wrong …

check it by differentiating, and you'll see how to fix it. ​

 

[darryw]

i knew it! this is an ongoing problem for me.. I always have problem escaping the integration loop when i have something like ∫t cos t.. (or even worse: ∫e^t cos t

as i understand it, the idea is to integrate up to a certain point and then subtract the integrals identity from left hand side, so then you cancel the integrals. When i did that i ended up with tsin t - t cos t. Can you offer any help/tips so i don't have to write out the whole long integration process? thanks

 

[tiny-tim]

i knew it! this is an ongoing problem for me.. I always have problem escaping the integration loop …

he he

the official method for ∫ fg dx is to integrate g only, giving [f(∫ g dx)], and then subtract the integral of f' time that: ∫ { f'(∫ g dx)} dx

but my method (only works for easy cases) is to make a guess (in this case, -tcost), differentiate it (-cost + tsint), and then integrate whatever's over (-cost)