Separable Differential Equations

So I have been getting caught up on two separable differential equations the first being:

$\displaystyle \frac {3du} {dt} = u^2$ with initial conditions: $\displaystyle u(0) = 4$

I divided by $\displaystyle u^2$ and multiplied by $\displaystyle dt$ giving:

$\displaystyle \frac {3du} {u^2} = dt$ and after integrating I got:

$\displaystyle \frac {3} {u} +C = t + C$ and inversing it, so that the u was in the numerator and multiplying by 3 I ended up with:

$\displaystyle u = \frac {3} {t+c}$

But I am not sure if this is how the problem is really supposed to be done.

The other equation I am having trouble with is finding the general term for:

$\displaystyle \frac {dR} {dx} = a(R^2+25)$

where a is some nonzero constant. Would this just be $\displaystyle 5tan(ax)+C$?

Thanks ahead of time for the help, I really appreciate it.

Quote:

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Originally Posted by Spudwad

So I have been getting caught up on two separable differential equations the first being:

$\displaystyle \frac {3du} {dt} = u^2$ with initial conditions: $\displaystyle u(0) = 4$

I divided by $\displaystyle u^2$ and multiplied by $\displaystyle dt$ giving:

$\displaystyle \frac {3du} {u^2} = dt$ and after integrating I got:

$\displaystyle \frac {3} {u} +C = t + C$ and inversing it, so that the u was in the numerator and multiplying by 3 I ended up with:

$\displaystyle u = \frac {3} {t+c}$

But I am not sure if this is how the problem is really supposed to be done.

The other equation I am having trouble with is finding the general term for:

$\displaystyle \frac {dR} {dx} = a(R^2+25)$

where a is some nonzero constant. Would this just be $\displaystyle 5tan(ax)+C$?

Thanks ahead of time for the help, I really appreciate it.

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$\displaystyle 3\frac{du}{dt} = u^2$

$\displaystyle 3u^{-2} du = dt$

Integrate:

$\displaystyle -3u^{-1} + K = t + K$

$\displaystyle \frac{-3}{u} = t + C$

$\displaystyle u = -\frac{3}{t+C}$

$\displaystyle u(0) = 4 \implies -\frac{3}{C}=4 \implies C = -\frac{3}{4}$

$\displaystyle u(t) = -\frac{3}{t-\frac{3}{4}}$