MHB DE 2.1.1.16 Find the solution of the give initial value problem

[karush]

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]

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

check the original DE statement again ...

 

[karush]

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

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]

that’s not what I meant ...

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

 

[karush]

that’s not what I meant ...

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

#16View attachment 9702

 

[Prove It]

#16

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]

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 ?

 

[skeeter]

$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

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]

$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

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??

 

[skeeter]

$\displaystyle yt^2 = \int \cos{t} \, dt$

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

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

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

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

 

[karush]

Ok saw my error