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Consider the initial value problem (IVP):

$$x'-\frac{x}{t}=t\cos(t),\ x(1)=-1$$

...and the possible solution:

$$x(t)=t^2\sin(t)-t$$

Choose the BEST answer.

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

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

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

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