Solution of the Differential equation $ u'(t)=4u^{3/4}(t) $ Let u(t) be a continuously differentiable function taking nonnegative values for $t\gt 0$ and satisfying $ u'(t)=4u^{3/4}(t) $; $ u(0)=0.\; Then \\ 1. u(t)=0 \\2. u(t)=t^4. \\3. u(t)= \begin{cases} 0 \qquad \qquad for \;\;0\lt t\lt 1 \\ (t-1)^4 \qquad for \;\;t\ge 1. \end{cases}\\4. u(t)=\begin{cases} 0 \qquad \qquad  for \;\;0\lt t\lt 10 \\ (t-10)^4 \qquad for \;\;t\ge 10. \end{cases}  $
My Attempt: given equation $\frac{du}{dt}=4u^{3/4} \\$
By variable seperable method,
$\frac{du}{4u^{3/4}}=dt $
On integrating we get 
$ u^{1/4}=t
\Rightarrow u=t^4 $ which gives option 2.
But i am not getting the other three solutions.
given that all the options are correct. thanks in advance.
 A: You seem to work under the assumption that ODEs have unique solutions. This is wrong is general. One theorem on uniqueness of solutions $\dot x = f(t,x)$, $x(t_0)=x_0$, is Picard-Lindelöf which needs local Lipschitz continuity of $f$ with respect to $x$. Your ODE is one of these examples where this is not fulfilled and indeed (as you can check by differentiating all four function that you have) the solutions stop to be unique as soon as the reach the value $x=0$ (which is the only point where $f$ is not locally Lipschitz).
A: Any function $u(t)=\max(0,t-c)^4$, $c\ge0$ is a solution, as it is a continuous differentiable function and its derivative $u'(t)=4\max(0,t-c)^3$ satisfies the differential equation. Your cases are the solutions for $c=\infty,0,1,10$.
A: Instead of attempting to solve the equation, you can plug the proposed solutions.


*

*Is of course valid, $(0)'=0^4\land 0=0$.

*and 3. and 4. $((t-c)^4)'=4((t-c)^4)^{3/4}$ for $t\ge c$ (and we already know that $0$ is fine elsewhere).
All these solutions are valid.
A: Only $u(t)=t^{4}$ and $u(t)=0$ are solutions which satisfy both the initial condition and the condition to be continuously differentiable! All other functions satisfy the equation and the initial condition, however, they are not continuously differentiable.
A: The integration should give
$u^{1/4}=t+c$ for any real $c$.
At $t=c$, $((t-c)^4)'=4(t-c)^3$
is zero so this is continuously differentiable.
A: A more rigorous resolution:
From the given equation, we can infer that $u(t)\ge0$. Then provided $u(t)>0$, we may separate and
$$\frac14u^{-3/4}du=dt$$ gives
$$u^{1/4}=t+c,$$
$$u(t)=(t+c)^4.$$
With the initial condition $u(0)=0$, we have $c^4=0$ and the only solution seems to be $u(t)=t^4.$

But, the option $u(t)=0$ remains, and is also a solution.
What's more, we can have $u(t)=0$ in an interval $[0,c]$, $c>0$, then switch to $(t-c)^4$, without losing differentiability. It is not possible to switch back to $0$ later, because then $u>0$.
