Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 966: 85


$P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$

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 $P(L, K_{0})$=$C_{1}(K_{0})\cdot L^{\alpha}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.