Answer
$P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
Work Step by Step
The separable differential equation is: $ \dfrac{dP}{dL}=\alpha\frac{P}{L}$
$ \int\dfrac{dP}{P}=\alpha \times \int \dfrac{dL}{L}$
$\ln|P|=\alpha\ln|L|+C$
Here, we have $C=C(K_{0})$ depends on $K_{0}$.
$|P|=e^{\alpha\ln|L|+C(K_{0})}=e^{\alpha\ln L} \times e^{C(K_{0})}=e^{\ln L^{\alpha}}\times e^{C(K_{0})}$
Thus, we get $|P|=L^{\alpha}\cdot C_{1}(K_{0});$ (another constant that depends on $K_{0}$)
Therefore, we get $P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
