You have a single node at the junction of the current source, the capacitor and the inductor. The sum of the currents into and out of the node must be identically 0.
The current in the inductor L must be the same current as in the resistor R. In any case, the voltage across the resistor must R times the inductor current.
That's two hints for the price of one.
Does that help?
actualy i found 4 equations but i didnt know if i missed something or didnt work right with them,
the equatins are :
i(t)=i_C+i_L
V_C=V_R+V_L
i_L'(t)*L=V_L
i_C(t)=C*V_L'
actualy i found 4 equations but i didnt know if i missed something or didnt work right with them,
the equatins are :
i(t)=i_C+i_L
V_C=V_R+V_L
i_L'(t)*L=V_L
i_C(t)=C*V_L'
I assume that by V_C you mean the voltage across the cap and by V_L you mean the voltage across the inductor, and by i_L'(t) you mean diL/dt, but what then is V_L' that looks like it would be dvL/dt. Is that what you really meant or did you mean to write v_C' because that would make more sense because you should see both derivatives in the expressions and then you should be able to solve this.
The two basic relationships are:
v=L*di/dt where i is iL, the current through the inductor and v is the voltage across the inductor vL, and
i=C*dv/dt where v is vC the voltage across the cap and i is the current though the cap iC..
For notation you can skip the (t) and the underscores like:
iL is current through inductor,
vC is voltage across cap,
i is the input current i(t).
This makes the results a bit more readable when using pure text.
Also, are you trying to write ODE's or just one 2nd order DE?