differential equation

**akhayoon** · April 21st 2008, 05:56 PM

$y\prime+\frac{2}{t}y=\frac{arctant}{t^{2}}$

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???

**Mathstud28** · April 21st 2008, 05:59 PM

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 $e^{\int\frac{2}{t}dt}$ ?