Answer:
A.x(t) = [tex]\frac{1}{C-kt}[\tex] satisfies the differential equation
Step-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} }[/tex]
k.x²=[tex]k.({\frac{1}{C-kt}})^{2} ={\frac{k}{(C-kt)^{2}}[/tex]
