Very hard differential equations help

Here are the 2 problems:
http://s4.postimage.org/40ecd8bwo/last2m…

Please help me. Do not worry about the direction field just help me solve those 2 problems. Include all the steps please because this will be on the exam.

Thank you so much

1)

y ' = -(2t + y)/2y

=> 2y y ' = - 2t - y ==> 2y y ' + y = -2t

=> y(2y ' + 1) = -2t

=> (y/t)(2y ' + 1) = -2

let y/t = v
y = vt
y ' = v + tv '

=> v(2v + 2tv ' + 1) = -2

=> 2v + 2tv ' + 1 = -2/v

=> 2tv ' = (-2/v) - 2v - 1

=> 2tv ' = - (2v^2 + v + 2)/v

separating variables

[ 2v dv /(2v^2 + v + 2) = - dt/t

=> [ v dv /(v^2 + v/2 + 1) = -dt/t

=> (1/2)[2v + 1/2 - 1/2 ] dv / (v^2 + v/2 + 1) = -dt/t

= (1/2)(2v + 1/2) dv / (v^2 + v/2 + 1) - (1/4) dv / (v^2 + v/2 + 1)= -dt/t

integrating

= (1/2) ln I v^2 + v/2 + 1) - 1/4∫ dv / [(v + 1/4)^2 + 15/16] = - ln(t) + c

= (1/2) ln I v^2 + v/2 + 1) - (√15 /15) tan^-1[v + 1/4] = - ln(t) + c

back substitute v = y/t

= (1/2) ln I (y/t)^2 + y/2t + 1) - (√15 /15) tan^-1[y/t + 1/4] = - ln(t) + c

= 1/2 ln I (1/2t^2)(2y^2 + yt + t^2) I - (√15 /15) tan^-1[(1/4t)(4y + t)] + ln(t) = C

= ln I SQRT(1/2t^2)(2y^2 + yt + t^2) I + ln(t) - (√15 /15) tan^-1[(1/4t)(4y + t)] = C

= ln I (1/t)SQRT((1/2)(2y^2 + yt + t^2) I + ln(t) - (√15 /15) tan^-1[(1/4t)(4y + t)] = C

= ln ISQRT((1/2)(2y^2 + yt + t^2) I - (√15 /15) tan^-1[(1/4t)(4y + t)] = C

= (1/2) ln I(1/2)(2y^2 + yt + t^2) I - (√15 /15) tan^-1[(1/4t)(4y + t)] = C


Since these problems are too lengthy, post the next problem separately.