Posted by **seth** on Saturday, March 19, 2011 at 5:36pm.

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

- separatio of variables -
**bobpursley**, Saturday, March 19, 2011 at 5:39pm
ydy=6x^2 dx

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

y(0)=4

16/2=C

- separatio of variables -
**Damon**, Saturday, March 19, 2011 at 6:00pm
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**, Saturday, March 19, 2011 at 8:17pm
damon is correct, I lost y.

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