The first is
Yes, the left side can be integrated using integration by parts.
The second is
with condition y(0)= -6
and the left side can be integrated using integration by parts.
Perhaps you have having trouble solving for y as a function of x? That normally is not required for problems like these but can be done. For example, integrating the second problem, you get (1/4)ln(|y- 2|)- (1/4)ln(|y+2|)= x+ C. Setting y= -6 when x= 0 gives (1/4)ln(8)- (1/4)ln(4)= (1/4)(ln(8)- ln(4))= (1/4)ln(8/4)= (1/4)ln(2)= C so we have (1/4)ln(|y-2|)- (1/4)ln(|y+2|)= x+ (1/4)ln(2). which we can write as ln|(y-2)/(y+2)|= 4x+ ln 2. Taking the exponential of both sides, |(y-2)/(y+2)|= = 2 . Since the initial value is at y= -6 the solution can only be valid for y< -2 which makes both numerator and denominator negative, so the fraction is positive, so we can remove the absolute values and solve for y.