Q:

Integrating Factor - Not getting the answer given?

$y' - 4y = t$ 
My integrating factor is $e^{-4t}$
$\int e^{-4t}y'$ - $\int 4e^{-4t}y$ = $\int te^{-4t}$
$\int (e^{-4t}y)'$ = $\int te^{-4t}$
$e^{-4t}y$ = $-4te^{-4t}$ - $e^{-4t}$ 
I end up with $y' = -4t -1 + ce^{4t}$
Where did I go wrong?

A:

You integrated $\int te^{-4t}\mathrm dt$ incorrectly.  It should be
$$\int te^{-4t}\mathrm dt=-\dfrac{t}{4}e^{-4t}-\dfrac{1}{16}e^{-4t}+c$$
Then multiplying both sides by $e^{4t}$ gives you 
$$y=-\dfrac{t}{4}-\dfrac{1}{16}+ce^{4t}$$

A:

Let $u=t$, $du=dt$, $dv=e^{-4t}$, and $v=-\cfrac{1}{4}e^{-4t}$, then using integration by parts, $\displaystyle\int te^{-4t}\, dt$ should be
\begin{align}
\int te^{-4t}\, dt&=-\cfrac{t}{4}e^{-4t}+\cfrac{1}{4}\int e^{-4t}\,dt\\
&=-\cfrac{t}{4}e^{-4t}+\cfrac{1}{4}\left(-\cfrac{1}{4}e^{-4t}\right)+C\\
&=-\cfrac{t}{4}e^{-4t}-\cfrac{1}{16}e^{-4t}+C\\
\end{align}

