Question

In a previous exercise we formulated a model for learning in the form of the differential equation dP dt = k(M − P) where P(t) measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P(t). (Use P for P(t). Assume that P(0) = 0.)

Answer 1

`(dP)/(M-P)= kdt`

And we can integrate both sides of the equation using the following substitution:

`u= M-P, du =-dP`

And replacing we got:

`\int -(du)/(u) = kt +C`

`-ln (u)= kt+c`

If we multiply both sides by -1 we got:

`ln (u ) = -kt -c`

`ln (M-P) = -kt -c`

And using exponential in both sides of the equation we got:

`M-P = e^(-kt) e^(-c)`

And solving for P we got:

`P(t) = M- e^(-kt)e^(-c)`

And replacing
`P_o =e^(-c)` we got:

`P(t) = M - P_o e^(-kt)`

We can use the condition
`P(0)=0` and we got:

`0 = M -P_o e^0`

And we see that
`M = P_o` and replacing we got:

`P= M(1- e^(-kt))`

Explanation:

For this case we aasume the following differential equation:

`(dP)/(dt)= k(M-P)`

Is a separable differential equation so we can do the following procedure:

`(dP)/(M-P)= kdt`

And we can integrate both sides of the equation using the following substitution:

`u= M-P, du =-dP`

And replacing we got:

`\int -(du)/(u) = kt +C`

`-ln (u)= kt+c`

If we multiply both sides by -1 we got:

`ln (u ) = -kt -c`

`ln (M-P) = -kt -c`

And using exponential in both sides of the equation we got:

`M-P = e^(-kt) e^(-c)`

And solving for P we got:

`P(t) = M- e^(-kt)e^(-c)`

And replacing
`P_o =e^(-c)` we got:

`P(t) = M - P_o e^(-kt)`

We can use the condition
`P(0)=0` and we got:

`0 = M -P_o e^0`

And we see that
`M = P_o` and replacing we got:

`P= M(1- e^(-kt))`