Finding the voltage across a inductor in a charged RLC-circuit
Sorry, it is incorrect, but you just asked a yes or no question. 

short Answer

since at t=0, I=10V/100ohms, =0.1A and Vc=10V every part will have exactly 10V across it. Thus Y1=-10V. This should be intuitive with Ohm’s Law or KVL. (And the switch has 10V across it too at t=0 due to equal resistors.)

There is no instant change in current or Cap voltage. This is because the parts are ideal with no ESR or DCR.



For high values of series R into C. At t=0 The impedance of XL=XC at some frequency which just happens to be near 100 Ohms so the flyback is (almost) critically damped starting near or = -10V.

You can work it out by Laplace, impedance ratios, or time domain ( and other ways).

Don't forget the Ic(t)=CdVc(t)/dt and VL(t)=LdI /dt while I, dI/dt &amp; dt  are equal in all parts now, while 100+100 = 200 ohms in the loop.