-2.1.6 ty'- 2y =\sin{t}, t>0

[karush]

2000
Find the general solution of the given differential equation
$\displaystyle
ty^\prime - 2y =\sin{t}, \quad t>0\\$
Divide thru by $t$
$\displaystyle y^\prime - \frac{2}{t}y =\frac{\sin{t}}{t}$
Obtain $u(t)$
$\displaystyle u(t)=\exp\int -
\frac{2}{t} \, dx =e^{2\ln{t}}=t^{-2}\\$
Multiply thru with $t^{-2}$
$t^{-2}y^\prime - 2 t y= t^{-2}\sin{t}\\$
Simplify:
$(t^{-2}y)'= t^{-2}\sin{t}\\$
Integrate:
$\displaystyle t^2y=\int t^{-2}\sin{t} dt =
2t\sin(t)-(t^2-2)\cos(t)+c_1\\$
Answer from textbook
$y=\color{red}
{(c-t\cos t + \sin t )/t^2} \quad
y \to 0 \textit{ as } t \to \infty$

ok somewhere I am not approaching the bk ans

 

[MarkFL]

When you multiply bt the integrating factor, you should have:

$$t^{-2}y'-2t^{-3}y=t^{-3}\sin(t)$$

Continue from there and see if you get the desired result. :)

 

[MarkFL]

To follow up, we may write:

$$\frac{d}{dt}\left(t^{-2}y\right)=t^{-3}\sin(t)$$

This does not lead to the solution given by your textbook. Going by that, we obtain:

$$\frac{d}{dt}\left(t^{2}y\right)=t\sin(t)$$

$$t^2y'+2ty=t\sin(t)$$

$$ty'+2y=\sin(t)$$

It appears either you copied the original ODE incorrectly, or your book gave it incorrectly.

 

[karush]

let me check
I had to stop doing this for while

due to 412 class

 

[MarkFL]

let me check
I had to stop doing this for while

due to 412 class

I apologize for replying to this months later...I was just going through the site looking for loose ends I've left. :)

 

[karush]

The original problem is this:

$$\displaystyle ty^\prime + 2y =\sin{t}, \quad t>0$$

I had a negative sign

unfortunately happens once every century

the book answer was apparently correct

Gosh, here I thot I got put on your 86 list!

 

[MarkFL]

The original problem is this:

$$\displaystyle ty^\prime + 2y =\sin{t}, \quad t>0$$

I had a negative sign

unfortunately happens once every century

the book answer was apparently correct

Gosh, here I thot I got put on your 86 list!

Most of your more recent questions are from courses I never took, and so I would rather more qualified folks reply. :D

 

[karush]

ok well the 400+ club seems to be here, thankfully...