Question

Part A: if k is a constant, show that a general (one-parameter) solution of the differential equation dx/dt = kx^2 is given by x(t) = 1/(C-kt), where C is an arbitrary constant. Part B: Determine by inspection a solution of the initial value problem dx/dt = kx^2, x(0) = 0.

Answer 1

We demonstrated that x(t) = 1/(C-kt) is a solution to the differential equation dx/dt = kx^2 by taking the derivative and showing it matches the original equation. For the initial value problem where x(0) = 0, x(t) = 0 for all t is a solution.

Step-by-step explanation:

To show that x(t) = 1/(C-kt) is a solution to the differential equation dx/dt = kx2, we can take the derivative of x(t) with respect to t and substitute it back into the original differential equation.

1. Take the derivative of x(t) with respect to t, which gives us dx/dt = k/(C-kt)2.
2. Substitute x(t) into the right-hand side of the original differential equation to get kx2 which simplifies to k/(C-kt)2.
3. Since the derivative of x(t) and kx2 are equal, x(t) = 1/(C-kt) is indeed a solution to the differential equation.

For Part B, by inspection, if x(0) = 0, then C must be zero to satisfy the initial condition. Therefore, the solution to the initial value problem is x(t) = 0 for all t.

Answer 2

A.x(t) =

k.x²=
`k.({(1)/(C-kt)})^(2) ={(k)/((C-kt)^(2))`