Odd results solving this very simple first-order RC circuit
Laplace Transform:

Solving in the s-domain will be easier. Writing the current division formula,

$$i_c (s) = \frac{R}{R+1/Cs}\times I(s)$$
$$i_c (s) = \frac{1}{1+1/5s}\times \frac{10}{s} =\frac{10}{s+0.2}$$

Taking the Laplace inverse,

$$i_c(t) = 10e^{-0.2t}u(t)$$
$$\therefore i_c(0^+) = 10A$$

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Another view to solve this problem: Capacitor acts as a short circuit for high frequency currents. At t = 0, a sudden switching from 0 to 10A (high frequency) occurs and capacitor acts as short circuit and hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$. 

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Using continuity condition on the capacitor's voltage:

Capacitor at \$t = 0^+\$ can be replaced with a voltage source with voltage = \$v_c(t=0^-)\$. But \$v_c(t=0^-) = 0\$. Voltage source with voltage = 0 is short circuit. Hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$.