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

[jahrens]

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.

 

[MarkFL]

Okay, we are given the IVP:

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

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

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

Application of the FTOC yields:

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

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

Solving for $u$, there results:

\(\displaystyle u(t)=\frac{12}{2-3t}\)

 

[jahrens]

That's it! Thank you so much!