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

separatio of variables 
bobpursley
ydy=6x^2 dx
y^2/2=2x^3 + C
y(0)=4
16/2=C 
separatio of variables 
Damon
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) 
separatio of variables 
bobpursley
damon is correct, I lost y.
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