Given is the differential equation

$N(t)' = 2 * \sqrt{N(t)}$

We have to show that the constant function N(t) = 0 is a solution for the initial condition N(0) = 0, and that the function N(t) = $t^2$ is a solution again for the same initial condition.

So I would plug in N(t) or N'(t) respectively in the above formula - but how do I use the initial conditions?


Hint: the equation is separable.

We can write:

$$\dfrac{1}{\sqrt{N}}dN = 2 dt$$

We can now integrate each side and then solve for the constant from the initial condition.

  • $\begingroup$ If I integrate both sides, I arrive at 2*N^(1/2) = 2t + C, is that correct? $\endgroup$ – TestGuest Nov 12 '13 at 13:32
  • 1
    $\begingroup$ Dive both sides by $2$, then, square both sides to solve for $N(t) = ...$. Now, use the initial condition $N(0) = 0$ to solve for $C$. $\endgroup$ – Amzoti Nov 12 '13 at 13:33
  • $\begingroup$ Got it, thanks! $\endgroup$ – TestGuest Nov 12 '13 at 13:36
  • $\begingroup$ You are welcome! Regards $\endgroup$ – Amzoti Nov 12 '13 at 13:38

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