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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?
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[dy/dx=24x^3y^5\] \[\rightarrow dy/y^5=24x^3dx\] \[\int\limits_{?}^{?}dy/y^5=\int\limits_{?}^{?}24x^3dx\] \[\rightarrow y^4/4=6x^4 +c\] Then solve it given the initial conditions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome nadeem, thanks for the quick reply :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so once I have c equal to 24, I just solve the original equation for y?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or solve from y^4/4=6x^4+24?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0solve from y^4/4=6x^4+24 by plugging in y=1 and x=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, sorry for the confusion

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem at all, thanks for all your help! you made my night of homework a lot less frustrating! :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm working on differential equations as well.... so frustrating sounds just about right
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