Welcome to our community

Be a part of something great, join today!

[SOLVED] De 17 y'-2y=e^{2t} y(0)=2

karush

Well-known member
Jan 31, 2012
2,678
#18
$ ty'+2y=\sin t \quad y(\pi/2)=1 \quad t\ge 0$
$y'+\dfrac{2}{t}y=\dfrac{\sin t}{t}$
so
$u(x) = e^{\int 2/t dt}= e^{t^2/4}$
 

Attachments

Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,693
#18
$ ty'+2y=\sin t \quad y(\pi/2)=1 \quad t\ge 0$
$y'+\dfrac{2}{t}y=\dfrac{\sin t}{t}$
so
$u(x) = e^{\int 2/t dt}= e^{t^2/4}$
$u(x) = e^{\int (2/t) dt}= e^{2\ln t} = \left(e^{\ln t}\right)^2 = t^2$.
 

karush

Well-known member
Jan 31, 2012
2,678
$u(x) = e^{\int (2/t) dt}= e^{2\ln t} = \left(e^{\ln t}\right)^2 = t^2$.
$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ???
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,693
$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ???
Yes, keep going! (Yes) (Integrate both sides with respect to $t$.)
 

karush

Well-known member
Jan 31, 2012
2,678
Yes, keep going! (Yes) (Integrate both sides with respect to $t$.)

$(yt^2)'=t\sin t$
IBP gives
$yt^2=-t\cos (t)+\sin (t)+C$
isolate y and reduce
$y=-\dfrac{\cos (t)}{t}+\dfrac{\sin (t)}{t^2}+\dfrac{c}{t^2}$


if ok next $y(\pi/2)=1$

isn't $t^2$ going to be ????
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y\left(\dfrac{\pi}{2}\right) = 1$

$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$

solve for $C$
 

karush

Well-known member
Jan 31, 2012
2,678
$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y\left(\dfrac{\pi}{2}\right) = 1$

$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$

solve for $C$
$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$
$\pi^2=4+4C$
$C=\dfrac{\pi ^2-4}{4}\approx1.4674 $


wasn't expecting a decimal but..


not sure how you would ck this?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655
what's up with the decimal? keep the exact value for $C$

$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{\pi^2-4}{4t^2}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \dfrac{4}{\pi^2} + \dfrac{\pi^2-4}{4 \cdot \frac{\pi^2}{4}}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \dfrac{4}{\pi^2} + \dfrac{\pi^2-4}{\pi^2}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \cancel{\dfrac{4}{\pi^2}} + \dfrac{\pi^2}{\pi^2} - \cancel{\dfrac{4}{\pi^2}}$

$y\left(\dfrac{\pi}{2}\right) = 1$
 

karush

Well-known member
Jan 31, 2012
2,678
$+ \dfrac{\pi^2-4}{4t^2}$

Where did this come from?
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
655