P3 differential eqn

Find the general solution of the differential equation

dx/dt = kx^2

where k is a constant, expressing x in terms of t in your answer.

Gvien that x =1 when t = 0 and x = 2 when t = 1, find the value of t near which x becomes very large.

dx/dt = kx^2
(x^-2)dx/dt = k
Integrate with respect to t to give
integral x^-2 .dx = integral k. dt
-1/x = kt + c

t=0 x=1 gives -1=c
so 1-1/x=kt
x=2 t=1 gives 1-1/2=k hence k=1/2
1-1/x=t/2
2(1-1/x)=t
when x gets large 1/x gets small, so t will be 2

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