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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?

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  1. anonymous
    • 5 years ago
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    \[dy/dx=24x^3y^5\] \[\rightarrow dy/y^5=24x^3dx\] \[\int\limits_{?}^{?}dy/y^5=\int\limits_{?}^{?}24x^3dx\] \[\rightarrow -y^4/4=6x^4 +c\] Then solve it given the initial conditions

  2. anonymous
    • 5 years ago
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    awesome nadeem, thanks for the quick reply :D

  3. anonymous
    • 5 years ago
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    no problem

  4. anonymous
    • 5 years ago
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    so once I have c equal to 24, I just solve the original equation for y?

  5. anonymous
    • 5 years ago
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    or solve from -y^4/4=6x^4+24?

  6. anonymous
    • 5 years ago
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    solve from -y^4/4=6x^4+24 by plugging in y=1 and x=-1

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    yeah, sorry for the confusion

  9. anonymous
    • 5 years ago
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    no problem at all, thanks for all your help! you made my night of homework a lot less frustrating! :D

  10. anonymous
    • 5 years ago
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    I'm working on differential equations as well.... so frustrating sounds just about right

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