The question is:

dy/dx = 12yx^3, y(0) = 6

My work:

(My work is attached) but here is the final answer I gety = 5e^(3x^4)which is supposedly wrong.

Any help would be greatly appreciated!

Thanks in advance!

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- Mar 23rd 2010, 02:15 PM #1

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- Mar 23rd 2010, 02:22 PM #2
Separate the variables

$\displaystyle \frac{dy}{y} = 12x^3\,dx$

$\displaystyle \ln |y| = 3x^4 + C$

$\displaystyle |y| = e^{3x^4+C} = Ae^{3x^4} \text{ where } A = e^C$

Initial condition: (0,6)

$\displaystyle 6 = Ae^0 \: \rightarrow \: A = 6$

Therefore the equation is $\displaystyle |y| = 6e^{3x^4}$