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solve differential equation dy/dt=-4y^3 y(0)=2.5
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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)
Why is 2C not absorbed to C?
I could do that, i would still get the same answer.
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
that's when you absorb -2 into the C
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