$(4+t^2)\frac{dy}{dt} + 2ty = 4t$
$(4+t^2)\frac{dy}{dt}= 4t -2ty$
$(4+t^2)dy = 4t dt -2tydt$
$\int (4+t^2)dy = \int( 4t -2ty )dt$
$4y+t^2y = 2t^2 -t^2y+ c$
$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}
Simple way is $$(4+t^2)\frac{dy}{dt} + 2ty = 4t$$ $$(4+t^2)y' + (4+t^2)'y = 4t$$ $$[(4+t^2)y]' = 4t$$ $$d[(4+t^2)y]= 4tdt$$ $$(4+t^2)y= 2t^2+C$$
your DE is not separable, that is why your answer is wrong from line 4.
If you don't realize that the left hand side is a product rule as it is posted in the solution, maybe you might try to write down the equation as
$$\frac{dy}{dt} + \frac{2t}{4+t^2}y = \frac{4t}{4+t^2}$$
and use a standard method to solve linear differential equations.
For solving matters, you are allowed "to multiply" by a differential ($dy,dt...$), but strictly speaking, these expressions are not just variables or numbers that you can manipulate freely algebraically, you need to be careful with what they truly mean and avoid these kind of common mistakes that you just committed.