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anonymous

  • 5 years ago

solve differential equation dy/dt=-4y^3 y(0)=2.5

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  1. anonymous
    • 5 years ago
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    separable differential equation, dy/y^3 = -4 dt integrate bothside -1/(2y^2) = -4t + c algebra. y(t)^2 = 1/(8t-2c) use initial condition y(0)=2.5 2.5^2 =1/(-2c) 25/4 = 1/ -2c -2/25 = c so your solution is y(t)^2 = 1/ (8t + 4/25) you can move the square around and make the solution y(t) = +/- 1/sqrt(8t + 4/25)

  2. anonymous
    • 5 years ago
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    Why is 2C not absorbed to C?

  3. anonymous
    • 5 years ago
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    I could do that, i would still get the same answer.

  4. anonymous
    • 5 years ago
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    y(t)^2 = 1/(8t+c) use initial condition y(0)=2.5 2.5^2 =1/(c) 25/4 = 1/ c 4/25 = c

  5. anonymous
    • 5 years ago
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    that's when you absorb -2 into the C

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