Question

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

A. P= -1+sqrt20t+25
B. P=5-e^-t
C. P=10-6e^-t
D. P=10-6e^t

Answer 1

• Option C.
  `P=10-6e^(-t)`

Step-by-step explanation:

1. Separate variables:

`(dP)/(dt)+P=10\\ \\ \\ \\(dP)/(dt)=-P+10\\ \\ \\ (dP)/(-P+10)=dt`

2. Find the indefinite integrals

`-ln((10-P))=t+C`

3. Solve for P

`10-P=e^(-t+C)\\ \\ 10-P=Ke^(-t)\\ \\ P=10-Ke^(-t)`

4. Use the initial condition, P(0) = 4, to find K:

• 4 = 10 - K(e⁰)

• 4 = 10 - K

• K = 10 - 4

• K = 6

5. Substitute K = 6 into the integrated equation:

`P=10-6e^(-t)`

That is the option C.