## anonymous 5 years ago I have a separable differential equation where (dy/dx)=24x^3y^5, and the initial value is at (-1, 1). Is the right start (-1/5)ln(y^5)= -30x^4+C?

1. anonymous

$dy/dx=24x^3y^5$ $\rightarrow dy/y^5=24x^3dx$ $\int\limits_{?}^{?}dy/y^5=\int\limits_{?}^{?}24x^3dx$ $\rightarrow -y^4/4=6x^4 +c$ Then solve it given the initial conditions

2. anonymous

3. anonymous

no problem

4. anonymous

so once I have c equal to 24, I just solve the original equation for y?

5. anonymous

or solve from -y^4/4=6x^4+24?

6. anonymous

solve from -y^4/4=6x^4+24 by plugging in y=1 and x=-1

7. anonymous

I thought plugging in the initial values gave me a method to solve c? but once I solve c and have to solve the equation for y do I solve from -y^4/4=6x^4+24?

8. anonymous

yeah, sorry for the confusion

9. anonymous

no problem at all, thanks for all your help! you made my night of homework a lot less frustrating! :D

10. anonymous

I'm working on differential equations as well.... so frustrating sounds just about right