MHB De 17 y'-2y=e^{2t} y(0)=2

[karush]

#18
$ ty'+2y=\sin t \quad y(\pi/2)=1 \quad t\ge 0$
$y'+\dfrac{2}{t}y=\dfrac{\sin t}{t}$
so
$u(x) = e^{\int 2/t dt}= e^{t^2/4}$

 

[Opalg]

#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]

$u(x) = e^{\int (2/t) dt}= e^{2\ln t} = \left(e^{\ln t}\right)^2 = t^2$.

$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ?

 

[Opalg]

$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ?

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

 

[karush]

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 ?

 

[skeeter]

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

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

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

solve for $C$

 

[karush]

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

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

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

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]

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]

$+ \dfrac{\pi^2-4}{4t^2}$

Where did this come from?

 

[skeeter]

$+ \dfrac{\pi^2-4}{4t^2}$

Where did this come from?

it’s $\dfrac{C}{t^2}$