# [SOLVED]2.1.6 Another DE went off the rails

#### karush

##### Well-known member
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:
$\displaystyle t^2y=\int t^{-2}\sin{t} dt = 2t\sin(t)-(t^2-2)\cos(t)+c_1\\$
$y=\color{red} {(c-t\cos t + \sin t )/t^2} \quad y \to 0 \textit{ as } t \to \infty$

ok somewhere Im not approaching the bk ans

#### MarkFL

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

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

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

#### MarkFL

Staff member
To follow up, we may write:

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

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

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

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

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

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

#### karush

##### Well-known member
let me check
I had to stop doing this for while

due to 412 class

#### MarkFL

Staff member
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

##### Well-known member
The original problem is this:

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

unfortunately happens once every century

the book answer was apparently correct

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

#### MarkFL

Staff member
The original problem is this:

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