# differential equation

• Apr 21st 2008, 06:56 PM
akhayoon
differential equation
$y\prime+\frac{2}{t}y=\frac{arctant}{t^{2}}$

here they want the y(t) equation

so first what I tried doing was integrating t/2

so I got the factor was $e^{lnt^{2}}$ which equals to 2

so now I have to integrate

$\frac{d}{dt}2y=2\frac{arctant}{t^2}$

which in the end would turn out to be

$y(t)=\frac{tarctant-0.5ln(t^{2}+1)}{t^{2}}+C/2$

so I can't help but feel I did something wrong???
• Apr 21st 2008, 06:59 PM
Mathstud28
Quote:

Originally Posted by akhayoon
$y\prime+\frac{2}{t}y=\frac{arctant}{t^{2}}$

here they want the y(t) equation

so first what I tried doing was integrating t/2

so I got the factor was $e^{lnt^{2}}$ which equals to 2

so now I have to integrate

$\frac{d}{dt}2y=2\frac{arctant}{t^2}+C/2$

which in the end would turn out to be

$y(t)=\frac{tarctant-0.5ln(t^{2}+1)}{t^{2}}$

so I can't help but feel I did something wrong???

are you sure it isnt $e^{\int\frac{2}{t}dt}$?
• Apr 21st 2008, 07:04 PM
akhayoon
yeah but wouldn't that be $e^{2lnt}$

which means that it would equal $e^{lnt^{2}} which equals 2$???
• Apr 21st 2008, 07:07 PM
Mathstud28
Quote:

Originally Posted by akhayoon
yeah but wouldn't that be $e^{2lnt}$

which means that it would equal $e^{lnt^{2}} which equals 2$???

$e^{2\ln(t)}=e^{\ln(t^2)}=t^2$...remember that since $e^{u(x)}$ and $\ln(u(x))$ are inverse operations $e^{\ln(u(x))}=u(x)$