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

Answer:

A.x(t) = $\frac{1}{C-kt}[\tex] satisfies the differential equationStep-by-step explanation:To show that [tex]\frac{1}{C-kt}[\tex] is a solution on one hand we derive the function and on the other we square it and multiply the constant k.[tex]\frac{dx}{dt} = \frac{d}{dt}\frac{1}{C-kt} =\frac{(-1).(-k)}{(C-kt)^{2} }=\frac{k}{(C-kt)^{2} }$

k.x²=$k.({\frac{1}{C-kt}})^{2} ={\frac{k}{(C-kt)^{2}}$