solve the initial value problem by separation of variables dy/dx=6x^2y and y(0)=4

Answer 1

ydy=6x^2 dx

y^2/2=2x^3 + C

y(0)=4
16/2=C

Answer 2

dy/y = 6 x^2 dx
ln y = 2 x^3 + c

when x = 0, y = 4
ln 4 = c
so
ln y = 2 x^3 + ln 4

ln (y/4) = 2 x^3

or
y/4 = e^(2x^3)
y = 4 e^(2x^3)

Answer 3

damon is correct, I lost y.

Answer 4

Suppose day/dx=6x+4pick possible expression for Y