You have found out the initial conditions of the circuit. You can also find out the steady state values after the switch has been turned on which are 3 A and 30 V.
To solve circuits with inductors and capacitors you have to write the differential equations. You have to redraw the circuit at t=0+.
You may start with this:
30 V = R1·iL + (L·diL/dt) (very simple 1st order linear and with constant coefficients differential equation; iL = current through the inductor)
Likewise for the voltage of the capacitor, exactly the same.
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The method showed on top is the mathematical approach and in my opinion is the way to go because the method is remembered always meanwhile formulas are easy forgattables.
However, you can memorise for the exam (and if the teacher allows you) the general solution of an Ordinary Differential Equation with constant coefficients (this case) which saves you time and is as follows:
x(t)=Xf + (X0-Xf)·e-(t-t0)/tau
Xf=final value i.e. steady state value
X0=initial condition
t0=time at which the initial condition is found out (in this case is t0 = 0 s)
tau=time constant
In conclusion:
For the inductor is as follows: iL(t)=3 A + (1 A - 3 A) · e-(t-0)/L/R1 = 3-2·ebla bla
Capacitor is as follows: vC(t)=30 V + (10 V - 30 V)·ebla bla