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

[darryw]

Homework Statement

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

Homework 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 don't 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 I am 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?

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Mark44]

Homework Statement

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

Homework Equations

integrating factor is t^2

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

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

Mistake in the line above.

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 don't need to also divide c by t^2, because a constant divided by a constant is still just another constant.. right?

No, you are not correct. You need to divide c by t^2 -- t^2 is not a constant.

so .. with that assumption...

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

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

this where I am 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?

Because the diff. equation is undefined for t = 0.

 

[darryw]

thanks for reply.

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 I am not grasping how to apply that info to solve the equation. thanks for any more help

 

[Mark44]

thanks for reply.

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 I am not grasping how to apply that info to solve the equation. thanks for any more help

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 = ?

 

[darryw]

if t = 0 then i can cancel last term

y = sin(t) / t^2

that is the solution to the problem?

 

[Mark44]

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.

 

[HallsofIvy]

if t = 0 then i can cancel last term

y = sin(t) / t^2

that is the solution to the problem?

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.