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Find the differential equation for Vo

My attempt at a solution:
Node V1:

$$\frac{V_1-V_{in}}{R_1} + \frac{1}{L}\int_{0}^{t}(V_1-V_2) = 0$$

Node V2:

$$\frac{1}{L}\int_{0}^{t}(V_2-V_1)+C*\dot{V_2}+\frac{V_2}{R_2}=0$$

Am I doing this correctly? How would I solve for V2? V2 is equal to Vo, correct? If the input (Vin) is a square wave, how would I find the transient and forced responses (assuming I'm given numerical values for R1, R2, L1, C1)?

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  • \$\begingroup\$ To be complete you have missed the node between Vin and V1: (V0 - V1)/R1 = I_vin, and V0 = Vin \$\endgroup\$
    – loudnoises
    Oct 17 '17 at 14:12
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Your equations are correct. Differentiate node \$\small V_2\$ equation and obtain an expression for \$\small V_1\$. Substitute this expression into the node \$\small V_1\$ equation. This gives a 2nd order equation in \$\small V_2\$.

Thus:

$$\small\ddot{V_2} +\left(\dfrac{1}{R_2C}+\dfrac{R_1}{L}\right)\dot{V_2}+\left(\dfrac{1}{LC}+\dfrac{R_1}{R_2 LC} \right)V_2=\dfrac{V_{in}}{LC} $$

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  • \$\begingroup\$ I ended up with this. Is it correct? $$\ddot{V_2}+\frac{\dot{V_2}}{L}(\frac{R_1}{R_2}+1)+\frac{V_2}{LC}(\frac{R_1}{R_2}+1)=\frac{V_{in}R_1}{R_2LC}$$ \$\endgroup\$
    – John
    Nov 26 '16 at 18:14

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