Question

Question 2Which of the following is the solution to the differential equation dP/dt +P=10 with the initial condition P(0) = 4

Answer 1

$\begin{gathered} \frac{dP}{dt}+P^{\prime}=10 \\ \\ \frac{1}{10-P}\times P^{\prime}=1 \end{gathered}$

On the left I make the integral with respect to P and on the right the integral with respect to t

$-(\ln (10-P))=t+c$$\begin{gathered} \ln (10-P)=-t+c \\ \end{gathered}$

we use e to simplify Steps to constant

$\begin{gathered} e^{\ln (10-P)}=e^{(-t+c)} \\ 10-P=e^{-t+c} \\ 10-P=e^{-t}\times e^c \\ 10-P=e^{-t}\times c \\ 10-P=ce^{-t} \end{gathered}$

now, solve P

$P=-ce^{-t}+10$

to find c we use P(0)=4, so when we replcae t=0 the soltuion must be 4

$\begin{gathered} 4=-ce^{-0}+10 \\ 4=-c+10 \\ c=10-4 \\ c=6 \end{gathered}$

the complete function is

$P=-6e^{-t}+10$

or

$P=10-6e^{-t}$

so, the right option is C