# [SOLVED]DE 2.1.1.16 Find the solution of the give initial value problem

#### karush

##### Well-known member
Find the solution of the give initial value problem

$\displaystyle y^\prime - \frac{2}{t}y =\frac{\cos{t}}{t^2}; \quad y{(\pi)}=0, \quad t>0$

$u(t)=e^{2 \ln{t}}$

then
$\displaystyle e^{2\ln{t}}\, y^\prime - \frac{2e^{e^{2\ln{t}}}}{t}y = \frac{e^{2\ln{t}}\cos{t}}{t^2}$

not sure actually!

#### skeeter

##### Well-known member
MHB Math Helper
$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

check the original DE statement again ...

#### karush

##### Well-known member
$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

check the original DE statement again ...
thus....

$\displaystyle e^{-2\ln{t}}\, y^\prime - \frac{-2e^{e^{2\ln{t}}}}{t}y = \frac{e^{-2\ln{t}}\cos{t}}{t^2}$

#### skeeter

##### Well-known member
MHB Math Helper
that’s not what I meant ...

please double check the original DE from wherever you got it from, specifically the right side

#### karush

##### Well-known member
that’s not what I meant ...

please double check the original DE from wherever you got it from, specifically the right side
#16

#### Prove It

##### Well-known member
MHB Math Helper
First of all, please note you made a typo in your original DE, which is why Skeeter said your integrating factor is wrong.

With the correct equation as in your picture, your integrating factor is correct, but you will not be able to do the integration unless you simplify it. $\displaystyle \mathrm{e}^{2\ln{(t)}} = \mathrm{e}^{\ln{\left( t^2 \right) }} = t^2$. Then the integration will be doable...

#### karush

##### Well-known member
Find the solution of the give initial value problem

$\displaystyle y^\prime + \frac{2}{t}y =\frac{\cos{t}}{t^2}; \quad y{(\pi)}=0, \quad t\ge 0$

$u(t)=e^{\displaystyle\ln{t^2}}$

then
$\displaystyle e^{\ln{t^2}}\, y^\prime +\frac{e^{\ln{t^2}}}{t}y = \frac{e^{\ln{t^2}}\cos{t}}{t^2}$

$\displaystyle(y'\cdot e^{\ln{t^2}})'=\frac{e^{\ln{t^2}}\cos{t}}{t^2}=\cos{ t}$

proceed ???

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

multiply evert term by the integrating factor, $e^{\int 2/t \, dt} = e^{2\ln{t}} = t^2$ ...

$y’t^2 + 2ty = \cos{t}$

$(yt^2)’ = \cos{t}$

now proceed

#### karush

##### Well-known member
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

multiply evert term by the integrating factor, $e^{\int 2/t \, dt} = e^{2\ln{t}} = t^2$ ...

$y’t^2 + 2ty = \cos{t}$

$(yt^2)’ = \cos{t}$

now proceed
$y(t)=\dfrac{\sin{t}}{t^2}+\frac{c}{t^2}$
So if $y(\pi)=0$ then
$y(\pi)=\dfrac{\sin{\pi}}{\pi^2}+\dfrac{c}{\pi^2}=0$
$c=-1$

Really??

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
$\displaystyle yt^2 = \int \cos{t} \, dt$

$yt^2 = \sin{t} + C$

$y = \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y(\pi) = 0 \implies 0 = \dfrac{\sin(\pi)}{\pi^2} + \dfrac{C}{\pi^2} \implies C = 0$

$y = \dfrac{\sin{t}}{t^2}$

Ok saw my error