Question

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

Answer 1

Answer:

•    Option C.    $P=10-6e^{-t}$

Explanation:

1. Separate variables:

      $\dfrac{dP}{dt}+P=10\\ \\ \\ \\\dfrac{dP}{dt}=-P+10\\ \\ \\ \dfrac{dP}{-P+10}=dt$

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.