Calc question thank u!!!!

Question

Solve the separable differential equation for 

du/dt=e^2u+4t

Use the following initial condition: 

u=???

Answer 1

du/dt=e^2u+4t

du/dt=e^{2u *} e^4t

du/e^2u= e^4t dt

∫ e^-2u du= ∫ e^4t dt

-1/2 e^-2u = 1/4 e^4t +C

e^-2u= -1/2 e^4t +C

ln e^-2u= ln (-1/2 e^4t +C)

-2u = ln (-1/2 e^4t +C)

u=( ln (-1/2 e^4t +C)) /-2

u(0) =12= (ln(-1/2 +C)) /-2

e^-24= = -1/2 +C

C= 0.5

u=( ln (-1/2 e^4t +1/2)) /-2

i hope it was clear!

Answer 2

du/dt = e^(2u) * e^(4t)

du/e^(2u) = e^(4t) dt

e^(-2u) du = e^(4t) dt

(-1/2) e^(-2u) = (1/4) e^(4t) + c

e^(-2u) = (-1/2) e^(4t) -2c

(-2u) = ln [ (-1/2) e^(4t) -2c]

u = (-1/2) ln [ (-1/2) e^(4t) -2c]

u(0)=12

12 = (-1/2) ln [ (-1/2) e^(4*0) -2c]

12 = (-1/2) ln [ (-1/2) - 2c]

-24 = ln [ (-1/2) - 2c ]

exp(-24) = (1/2) - 2c

2c = (1/2) - exp(-24)

c = (1/4) - (1/2)exp(-24)