RLC Circuit, Find Diff Eq for VC - Can someone check my work?

[eehelp150]

Homework Statement

Find the differential equation for VC
Circuit #1

Circuit #2

Homework Equations

KCL

The Attempt at a Solution

Circuit #1:
Node VC:
$\frac{V_C-V_{in}}{R_1}+\frac{1}{L_1}\int_{0}^{t}(V_C-V_2)+C_1\dot{V_C}=0$
Node V2:
$\frac{1}{L_1}\int_{0}^{t}(V_2-V_C)+\frac{V_2}{R_2}=0$

Derive NodeVC and solve for V2
$\frac{\dot{V_C}}{R_1}-\frac{\dot{V_{in}}}{R_1}+\frac{V_C}{L_1}-\frac{V_2}{L_1}+C_1\ddot{V_C}=0$
$\frac{\dot{V_C}}{R_1}-\frac{\dot{V_{in}}}{R_1}+\frac{V_C}{L_1}+C_1\ddot{V_C}=\frac{V_2}{L_1}$
$\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}=V_2$

Plug that into node V2 equation
$\frac{1}{L_1}\int_{0}^{t}((\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C})-V_C)+\frac{\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}}=0$

Simplify
$\frac{1}{L_1}\int_{0}^{t}(\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+C_1L_1\ddot{V_C})+\frac{\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}}{R_2}=0$
get rid of integral
$\frac{1}{L_1}(\frac{L_1{V_C}}{R_1}-\frac{L_1{V_{in}}}{R_1}+C_1L_1\dot{V_C})+\frac{\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}}{R_2}=0$
Simplify
$(\frac{{V_C}}{R_1}-\frac{{V_{in}}}{R_1}+C_1\dot{V_C})+\frac{\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}}{R_2}=0$

Multiply everything by R2
$R_2(\frac{{V_C}}{R_1}-\frac{{V_{in}}}{R_1}+C_1\dot{V_C})+\frac{L_1\dot{V_C}}{R_1}-\frac{L_1\dot{V_{in}}}{R_1}+{V_C}+C_1L_1\ddot{V_C}=0$

I end up with:
$\ddot{V_C}+\frac{\dot{V_C}}{R_1C_1}+\frac{\dot{V_C}}{L_1}+\frac{V_CR_2}{R_1C_1L_1}+\frac{V_C}{C_1L_1}=\frac{V_{in}}{R_1C_1}+\frac{V_{in}}{R_1C_1L_1}$

Circuit #2 (V2=VC)
Node V1
$\frac{V_1-V_{in}}{R_1}+\frac{1}{L}\int_{0}^{t}(V_1)+\frac{V_1-V_2}{R_2}=0$
Node V2
$\frac{V_2-V_1}{R_2}+C\dot{V_2}=0$
Solve NodeV2 for V1
$V_2-V_1+R_2C\dot{V_2}=0$
$V_1=V_2+R_2C\dot{V_2}$
plug into NodeV1
$\frac{V_2+R_2C\dot{V_2}-V_{in}}{R_1}+\frac{1}{L}\int_{0}^{t}(V_2+R_2C\dot{V_2})+\frac{V_2+R_2C\dot{V_2}-V_2}{R_2}=0$

Simplify
$\frac{V_2}{R_1}+\frac{R_2C\dot{V_2}}{R_1}-\frac{V_{in}}{R_1}+\frac{1}{L}\int_{0}^{t}(V_2+R_2C\dot{V_2})+C\dot{V_2}=0$
Derive everything to get rid of integral
$\frac{\dot{V_2}}{R_1}+\frac{R_2C\ddot{V_2}}{R_1}-\frac{\dot{V_{in}}}{R_1}+\frac{V_2}{L}+\frac{R_2C\dot{V_2}}{L}+C\ddot{V_2}=0$

Combine like terms
$C\ddot{V_2}+\frac{R_2C\ddot{V_2}}{R_1}+\frac{\dot{V_2}}{R_1}+\frac{R_2C\dot{V_2}}{L}+\frac{V_2}{L}=\frac{\dot{V_{in}}}{R_1}$
Divide everything by C
$\ddot{V_2}+\frac{R_2\ddot{V_2}}{R_1}+\frac{\dot{V_2}}{CR_1}+\frac{R_2\dot{V_2}}{L}+\frac{V_2}{LC}=\frac{\dot{V_{in}}}{CR_1}$
Simplify
$\ddot{V_2}(1+\frac{R_2}{R_1})+\dot{V_2}(\frac{1}{CR_1}+\frac{R_2}{L})+\frac{V_2}{LC}=\frac{\dot{V_in}}{CR_1}$

Am I doing these two problems right?

 

[KataKoniK]

I cannot confirm whether your solutions are correct without seeing the original circuit diagrams and the specific values for the components. However, your approach seems reasonable and the mathematical steps look correct. I would recommend double-checking your work and verifying with the given values to ensure accuracy. Additionally, it would be helpful to label your equations and clearly state your assumptions and any simplifications made.