# Differential Equation dont understand t>0 how to apply that to equation

1. Apr 11, 2010

### darryw

1. The problem statement, all variables and given/known data

y' + (2/t)y = cos(t) / t^2 initial cond: y(pi) = 0 .... t>0

2. Relevant equations
integrating factor is t^2

so.. integ (yt^2)' = integ cos (t)

= yt^2 = -sin (t) + c

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

..this brings up my first question.. when i divided through by t^2 to isolate y, am i correct in saying that I dont need to also divide c by t^2, because a constant divided by a constant is still just another constant.. right?

so .. with that assumption...

0 = ( - sin (pi) / pi^2 )+ c

0 = ( - 0 / pi^2 ) + c
0 = c

this where im stuck, because i realize IC said that t>0.. but what does it mean to apply that IC to the equation?? Does this lead to an actual value for c somehow? thanks
edit: if it says that t = pi .. that is, y(pi) = 0 .. then why would it also say t>0 ?? pi is clearly bigger than zero, so why the added condition?

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 11, 2010

### Staff: Mentor

Mistake in the line above.
No, you are not correct. You need to divide c by t^2 -- t^2 is not a constant.
Because the diff. equation is undefined for t = 0.

3. Apr 11, 2010

### darryw

yt^2 = sin(t) + c

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

IC: y(pi) = 0

0 = 0 / pi^2 + c/pi^2

c = 0(pi^2)

c = 0

i understand why it is undefined at zero but Im not grasping how to apply that info to solve the equation. thanks for any more help

4. Apr 11, 2010

### Staff: Mentor

You don't need to apply the information that t > 0. And you almost there with the solution.

You have y = sin(t) / t^2 + c/t^2 and c = 0, so y = ?

5. Apr 11, 2010

### darryw

if t = 0 then i can cancel last term

y = sin(t) / t^2

that is the solution to the problem?

6. Apr 11, 2010

### Staff: Mentor

No, t can't be zero.

You have y(t) = sin(t) / t^2 + C/t^2 and y(pi) = 0, so
0 = y(pi) = sin(pi)/t^2 + C/t^2
==> C/t^2 = 0 ==> C = 0

So y(t) = sin(t) / t^2, and we still have the restriction that t > 0.

7. Apr 12, 2010

### HallsofIvy

Staff Emeritus
You mean "if c= 0". And you should include "t> 0" in your solution:

y= sin(t)/t^2, t> 0.

Since the differential equation does not hold at 0, not only can t not be 0 but you cannot continue the solution to negative values of t.