While I am solve $(4+t^2)\frac{dy}{dt} + 2ty = 4t$, I don't know why my answer is wrong.. $(4+t^2)\frac{dy}{dt} + 2ty = 4t$
$(4+t^2)\frac{dy}{dt}= 4t -2ty$
$(4+t^2)dy = 4t dt -2tydt$
$\int (4+t^2)dy = \int( 4t -2ty )dt$
$4y+t^2y = 2t^2 -t^2y+ c$
$y(4+2t^2)= 2t^2 +c $
$y = \frac{2t^2 + c}{4+2t^2}$
This is my answer, but the text book answer is $y = \frac{2t^2 + c}{4+t^2}$  
This is text book.... answer and solution 

I don't know why my answer is wrong.
so... to solve this kind of question... can't I solve like what I did ? ( multiply both side with dt)

The basic error was integrating before the variables were separated.
\begin{eqnarray}
(4+t^2)\frac{dy}{dt} + 2ty &=& 4t\\
(4+t^2)\,dy+2ty\,dt&=&4t\,dt\\
(4+t^2)\,dy+(2ty-4t)\,dt&=&0\\
(4+t^2)\,dy+2t(y-2)\,dt\\
\frac{1}{y-2}\,dy+\frac{2t}{4+t^2}\,dt&=&=0\\
\ln\vert y-2\vert+\ln\vert 4+t^2\vert&=&\ln\vert c\vert\\
\ln\vert(y-2)(4+t^2)\vert&=&\ln\vert c\vert\\
(y-2)(4+t^2)&=&c\\
y-2&=&\frac{c}{4+t^2}\\
y&=&\frac{c}{4+t^2}+2\\
y&=&\frac{c+8+2t^2}{4+t^2}\\
y&=&\frac{2t^2+C}{4+t^2}
\end{eqnarray}
