1. I.v.p ode

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?

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.

3. wow thanks a lot did not know that seperable can be used lol i forgottern about it

4. You should never forget about separability. If it's possible to separate variables, then chances are that separating is about the easiest way to solve a DE. That's your first line of defense.

5. Originally Posted by Ackbeet
You should never forget about separability. If it's possible to separate variables, then chances are that separating is about the easiest way to solve a DE. That's your first line of defense.
Not to mention that many of the other methods (e.g. integrating factor) are only possible because from analysis of the situation, they were able to find something seperable...