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# Solution of IVP - $x(t)=\frac{1}{1-t}$

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Consider the initial value problem (IVP):

$$x'=x^2,\ x(0)=2$$

...and the possible solution:

$$x(t)=\frac{1}{1-t}$$

A

The function $x$ is a solution of the IVP.

B

The function $x$ satisfies the initial condition, but not the differential equation.

C

The function $x$ satisfies the differential equation, but not the initial condition.

D

The function $x$ does not satisfy either the differential equation or the initial condition.