A special class of first-order linear equations have the form $a(t)y'(t) + a'(t)y(t) = f(t)$, where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $ rac{d}{dt}(a(t)y(t)) = a(t)y'(t) + a'(t)y(t) = f(t)$. Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following.
$ty'(t) + y = 4 + 2t$, $y(1) = 6$
y(t) = , where t > 0
