hi to all once again

how do i solve this equation

y'=4x^3.(y-2) subject to y(0)=-3

1)y(x)=2+5^(x^4)

2)y(x)=2-5^(x^4)

i having problem dy/dx 5^(x^4) how do i do it?using log?,if i use log do i have to log y and 2 too?

Thanks in advance

Results 1 to 5 of 5

- Apr 3rd 2011, 10:42 PM #1

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- Apr 3rd 2011, 11:28 PM #2
The ODE is seperable.

$\displaystyle \displaystyle \frac{dy}{dx} = 4x^3(y-2)$

$\displaystyle \displaystyle \frac{1}{y - 2}\,\frac{dy}{dx} = 4x^3$

$\displaystyle \displaystyle \int{\frac{1}{y - 2}\,\frac{dy}{dx}\,dx} = \int{4x^3\,dx}$

$\displaystyle \displaystyle \int{\frac{1}{y - 2}\,dy} = \int{4x^3\,dx}$.

Go from here.

- Apr 4th 2011, 12:32 AM #3

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- Apr 4th 2011, 02:12 AM #4

- Apr 4th 2011, 02:17 AM #5