Solve the Initial Value Problem:
Let k and Po be positive constants. P(0)=Po
dP/dt = kP^2
That looks like a separable first-order differential equation rather than calc 2, but here goes:
(separate both variables)
P^-2 dP = k dt
(integrate both sides)
-1/P = kt + C
P = -1 / (kt + C)
Po = -1 / C
C = -1 / Po
P = -1 / (kt + -1/Po)
To solve the initial value problem, we need to find the function P(t) that satisfies the given differential equation dP/dt = kP^2, with the initial condition P(0) = Po.
To solve the differential equation, we can separate variables by writing it in the form:
dP / P^2 = k dt
Now, let's integrate both sides:
∫ (1 / P^2) dP = ∫ k dt
The integral of (1 / P^2) is -1 / P, and the integral of k dt is kt. Applying the definite integral from 0 to t on both sides:
∫ (1 / P^2) dP from Po to P = ∫ k dt from 0 to t
Now, let's integrate:
[-1 / P] from Po to P = k(t - 0)
Substituting the limits of integration and simplifying:
[-1 / P] - (-1 / Po) = kt
Rearranging the equation:
1 / P - 1 / Po = kt
To find P, we need to isolate it. We can start by adding 1 / Po to both sides:
1 / P = 1 / Po + kt
Then, take the reciprocal of both sides:
P = 1 / (1 / Po + kt)
This is the solution to the initial value problem, given P(0) = Po.