I'm currently taking Physics II and I'm tasked to solve for the voltage $$U_R(t)$$ in this circuit.

It's driven by a current source $$I(t) = I_0*e^{i*\omega*t}$$ and I can neglect the switch on process.

I'm currently at the point of having the differential equation to solve this problem but I'm not entirely sure on how to solve it yet.

I wanted to ask if my thought process was correct so far and if I landed at the correct differential equation or am I on the completely wrong path?

## Best Answer

Yes, your analysis is correct.

It would have been quicker to solve for the resistor current and then multiply through by \$\small R\$ to get the resistor voltage, \$v_R\$, as below (I've used \$\small j=\sqrt{-1}\$ to avoid confusion with current, \$i)\$.

Let \$i\$ be the resistor current, then the voltage across the current source is: $$L\frac{d}{dt}(I_0e^{j\omega t}-i)=\frac{1}{C}\int i\:dt+Ri$$ differentiate,

$$L\frac{d^2i}{dt^2}+R\frac{di}{dt}+\frac{1}{C}i=L\frac{d^2}{dt^2}(I_0e^{j\omega t})$$

multiply by \$\frac{R}{L}\$,

$$\frac{d^2v_R }{dt^2}+\frac{R}{L}\frac{dv_R}{dt}+\frac{1}{LC}v_R=-R\omega^2I_0e^{j\omega t}$$