MHB How to Solve a Separable Equation with Initial Condition u(0)=6?

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4 du/dt = u^2 with initial condition u(0)=6

I have worked this multiple times, and all I get is u = (-8/(t-27))^(1/3) and it is NOT right! If anyone can help it would be very appreciated.
 
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Okay, we are given the IVP:

$$4\d{u}{t}=u^2$$ where $$u(0)=6$$

Separating variables, switching out the dummy variables of integration and using the boundaries, we obtain:

$$4\int_6^u v^{-2}\,dv=\int_0^t\,dw$$

Application of the FTOC yields:

$$-4\left[\frac{1}{v}\right]_6^u=[w]_0^t$$

$$4\left(\frac{1}{6}-\frac{1}{u}\right)=t$$

Solving for $u$, there results:

$$u(t)=\frac{12}{2-3t}$$
 
That's it! Thank you so much!
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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