Solving the Differential Equation $u'=u^{1-k}$ by separation of variables.

Question

I have the differential equation $u'=u^{1-k}$, and am supposed to solve it by separation of variables. As an attempted solution, I have $u^{k-1}du=dt$, and therefore $\frac{u^k}{k}=t+c$, and therefore $u=(kt)^\frac{1}{k}+c$. With $u(0)=1$, this is $u=(kt)^\frac{1}{k}+1$. Is this the correct solution, and also, for what values of $k<0$ does $u$ "blow up".

Answer 1

It's ok to absorb $k$ into $C$, but it's not ok to say $(kt+C)^{\frac{1}{k}}=(kt)^{\frac{1}{k}}+C$

The value for $C$ given $u(0)$ will turn out to be the same, but not for the same reason. (What if $u(0)\neq1$ or $u(0)\neq0$)

Answer 2

$$u'=u^{1-k}$$ you were almost there just you made a little mistake in the last step... $u=(kt)^\frac{1}{k}+c$

First for $k=0$ $$\int \frac {du}u=\int dt \implies \ln|u|=t+K \implies u=Ke^t$$ $$u(0)=1 \implies K=1 \implies u(t)=e^t$$ for $k \neq 0$ $$\int \frac {du}{u^{1-k}}=\int dt \implies u^k=kt+C \implies u=(kt+C)^{\frac1k}$$ $$u(0)=1 \implies C=1 \implies u(t)=(kt+1)^{\frac1k}$$