Integrating Factor error

**astuart** · November 9th 2012, 03:15 PM

Hello,
Could someone point out where I'm going wrong here - I'm making a mistake somewhere, and it's not clear to me what I'm doing wrong...

$\frac{dy}{dt}(t) - 3y +2 = 0$

$= \frac {dy}{dt} - \frac {3y}{t} = \frac {-2}{t}$

$I(t)=e^{\int \frac {-3}{t}}$

$=I(t) = e^{-3lnt} = t^{-3}$

$= \frac {dy}{dt}\cdot t^{-3} \frac {-3y}{t}\cdot t^{-3} = \frac {-2}{t} \cdot t^{-3}$

$= \int \frac {d}{dt}t^{-3}y = \int \frac {-2}{t^4}$

$= yt^{-3} = -2 \int t^{-4}$

$= yt^{-3} = -2 \frac {t^{-3}}{-3}+C$

$= yt^{-3} = \frac {2}{3} t^{-3}+C = \frac {2}{3t^3}+C$

$= y = \frac {2}{3} + C$

Now, I end up cancelling the t-value out, which can't be right. What am I doing wrong exactly?

**astuart** · November 9th 2012, 03:22 PM

Originally Posted by astuart

Hello,
Could someone point out where I'm going wrong here - I'm making a mistake somewhere, and it's not clear to me what I'm doing wrong...

$\frac{dy}{dt}(t) - 3y +2 = 0$

$= \frac {dy}{dt} - \frac {3y}{t} = \frac {-2}{t}$

$I(t)=e^{\int \frac {-3}{t}}$

$=I(t) = e^{-3lnt} = t^{-3}$

$= \frac {dy}{dt}\cdot t^{-3} \frac {-3y}{t}\cdot t^{-3} = \frac {-2}{t} \cdot t^{-3}$

$= \int \frac {d}{dt}t^{-3}y = \int \frac {-2}{t^4}$

$= yt^{-3} = -2 \int t^{-4}$

$= yt^{-3} = -2 \frac {t^{-3}}{-3}+C$

$= yt^{-3} = \frac {2}{3} t^{-3}+C = \frac {2}{3t^3}+C$

$= y = \frac {2}{3} + C$

Now, I end up cancelling the t-value out, which can't be right. What am I doing wrong exactly?

Ahh, god, I think I can already see where I went wrong....

$y = \frac {2}{3} + Ct^3$

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