The derivative of
\(A_1 t e^{-1500 t}\) is actually
\(A_1 e^{-1500 t} + A_1 t (-1500) e^{-1500 t}\)
so, you should correct that.
Also, for times before zero, the voltage is constant and as a result, the current of the capacitor is zero, not 90mA as you wrote.
Since the voltage on a capacitor is a continuous quantity, you can safely say that
\(A_1 \cdot 0 \cdot e^{-1500 \cdot 0}+A_2 \cdot e^{1500 \cdot 0}=A_2=25V\).
A typical (if not unique) circuit which would satisfy the general solution would be a capacitor with an initial voltage [25V] discharging into a series R+L network. The effective circuit would be critically damped.
Under these conditions one would obtain the general solution for the capacitor voltage
This solution supports Georacer's comments about the initial capacitor current being zero. If there were a current at zero, the general solution for the cap voltage may not be satisfied - I've not checked whether this is in fact the case. It's probably beyond the original scope of the problem anyway.