Originally Posted by

**Coop** Hi,

This is first post, will I be able to post in the normal section after this?

EDIT: ^Ignore :P I figured it out.

Anyway, one of my homework problems reads like this

"*Verify that y = -tcos(t)-t is a solution to the initial value problem *

t(dy/dt) = y + sin(t); y(pi)=0"

I solved it,

*dy/dt = (-tcos(t)-t)' = t(sin(t))-cos(t)-1*

After plugging y and y' in,

t(tsin(t) - cos(t) - 1) = (-tcos(t) - t) + t^{2}sin(t)

Simplifying,

t^{2}sin(t) - tcos(t) - t = t^{2}sin(t) - tcos(t) - t

And,

-tcos(t) - t = -tcos(t) - t

But my question is, why did they give me the information that y(pi) = 0? Although it's true, am I supposed to use it anywhere? Is it just extra info. to throw me off?