Question

\[ \begin{array}{l} \int \frac{d P}{P}=\int k d t \\ \ln |P|=k t+c \\ \ln P_{0}=k t+c \\ \ln P=k t+\ln P_{0} \\ \ln P-\ln P_{0}=k t \\ \ln P \mid P_{0}=k t \\ \frac{P}{P_{0}}=e^{k t} \\ P=P_{0} e^{k t} \end{array} \] Explain the obore

          \[
\begin{array}{l}
\int \frac{d P}{P}=\int k d t \\
\ln |P|=k t+c \\
\ln P_{0}=k t+c \\
\ln P=k t+\ln P_{0} \\
\ln P-\ln P_{0}=k t \\
\ln P \mid P_{0}=k t \\
\frac{P}{P_{0}}=e^{k t} \\
P=P_{0} e^{k t}
\end{array}
\]

Explain the obore

∫(d P)/(P)=∫ k d t
ln |P|=k t+c
ln P0=k t+c
ln P=k t+ln P0
ln P-ln P0=k t
ln P | P0=k t
(P)/(P0)=e^k t

P=P0 e^k t

Explain the obore

Added by Purificaci-N C.

Calculus

Deborah Hughes-Hallett, Andrew M.… 5th Edition

Chapter 1

Instant Answer

Solved by Expert William Mead

Step 1

Step 1: Start with the given differential equation: \[ \int \frac{dP}{P} = \int k \, dt \] Show more…

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\[ \begin{array}{l} \int \frac{d P}{P}=\int k d t \\ \ln |P|=k t+c \\ \ln P_{0}=k t+c \\ \ln P=k t+\ln P_{0} \\ \ln P-\ln P_{0}=k t \\ \ln P \mid P_{0}=k t \\ \frac{P}{P_{0}}=e^{k t} \\ P=P_{0} e^{k t} \end{array} \] Explain the obore