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

karush

Well-known member
Jan 31, 2012
2,678
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
Mar 1, 2012
655
$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

check the original DE statement again ...
 

karush

Well-known member
Jan 31, 2012
2,678
$e^{\int -\frac{2}{t} \, dt} = e^{-2\ln{t}} = \dfrac{1}{t^2}$

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
Mar 1, 2012
655
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
Jan 31, 2012
2,678
that’s not what I meant ...

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


2020_04_23_18.52.40.jpg
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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
Jan 31, 2012
2,678
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
Mar 1, 2012
655
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

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
Jan 31, 2012
2,678
$y’ + \dfrac{2}{t} \cdot y = \dfrac{\cos{t}}{t^2}$

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
Mar 1, 2012
655
$\displaystyle yt^2 = \int \cos{t} \, dt$

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

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

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

$y = \dfrac{\sin{t}}{t^2}$
 

karush

Well-known member
Jan 31, 2012
2,678
Ok saw my error