contestada

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

Respuesta :

Edufirst

Answer:

  •    Option C.    [tex]P=10-6e^{-t}[/tex]

Explanation:

1. Separate variables:

      [tex]\dfrac{dP}{dt}+P=10\\ \\ \\ \\\dfrac{dP}{dt}=-P+10\\ \\ \\ \dfrac{dP}{-P+10}=dt[/tex]

2. Find the indefinite integrals

    [tex]-\ln{(10-P)}=t+C[/tex]

3. Solve for P

      [tex]10-P=e^{-t+C}\\ \\ 10-P=Ke^{-t}\\ \\ P=10-Ke^{-t}[/tex]

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:

       [tex]P=10-6e^{-t}[/tex]

That is the option C.

Otras preguntas