Respuesta :
Answer:
[tex] \frac{dP}{M-P}= kdt[/tex]
And we can integrate both sides of the equation using the following substitution:
[tex] u= M-P, du =-dP[/tex]
And replacing we got:
[tex] \int -\frac{du}{u} = kt +C[/tex]
[tex] -ln (u)= kt+c[/tex]
If we multiply both sides by -1 we got:
[tex] ln (u ) = -kt -c[/tex]
[tex] ln (M-P) = -kt -c[/tex]
And using exponential in both sides of the equation we got:
[tex] M-P = e^{-kt} e^{-c}[/tex]
And solving for P we got:
[tex] P(t) = M- e^{-kt}e^{-c}[/tex]
And replacing [tex] P_o =e^{-c}[/tex] we got:
[tex] P(t) = M - P_o e^{-kt}[/tex]
We can use the condition [tex] P(0)=0[/tex] and we got:
[tex] 0 = M -P_o e^0 [/tex]
And we see that [tex] M = P_o[/tex] and replacing we got:
[tex] P= M(1- e^{-kt})[/tex]
Step-by-step explanation:
For this case we aasume the following differential equation:
[tex] \frac{dP}{dt}= k(M-P)[/tex]
Is a separable differential equation so we can do the following procedure:
[tex] \frac{dP}{M-P}= kdt[/tex]
And we can integrate both sides of the equation using the following substitution:
[tex] u= M-P, du =-dP[/tex]
And replacing we got:
[tex] \int -\frac{du}{u} = kt +C[/tex]
[tex] -ln (u)= kt+c[/tex]
If we multiply both sides by -1 we got:
[tex] ln (u ) = -kt -c[/tex]
[tex] ln (M-P) = -kt -c[/tex]
And using exponential in both sides of the equation we got:
[tex] M-P = e^{-kt} e^{-c}[/tex]
And solving for P we got:
[tex] P(t) = M- e^{-kt}e^{-c}[/tex]
And replacing [tex] P_o =e^{-c}[/tex] we got:
[tex] P(t) = M - P_o e^{-kt}[/tex]
We can use the condition [tex] P(0)=0[/tex] and we got:
[tex] 0 = M -P_o e^0 [/tex]
And we see that [tex] M = P_o[/tex] and replacing we got:
[tex] P= M(1- e^{-kt})[/tex]
