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I have an RC circuit as shown below enter image description here

unfortunately I have no knowledge about RC solving but I want to obtain corresponding differential equations for V1, V2 and V3. Having in hand differential equation for V1 is enough for now, if anyone can tell me it!!!

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    \$\begingroup\$ s-domain analysis is one way to go. \$\endgroup\$ – K. Rmth Nov 7 '15 at 17:32
  • \$\begingroup\$ I'll be grateful if you can give me differential equation for only V1 \$\endgroup\$ – Mahmoud Nov 7 '15 at 18:04
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    \$\begingroup\$ I'm voting to close this question as off-topic because homework without effort. \$\endgroup\$ – Brian Carlton Nov 7 '15 at 22:46
  • \$\begingroup\$ it's not homework. actually it is a circuit model from a paper which I need to use it for my analysis in my dissertion \$\endgroup\$ – Mahmoud Nov 8 '15 at 6:34
  • \$\begingroup\$ Possible duplicate of Deriving 2nd order passive low pass filter cutoff frequency \$\endgroup\$ – jippie Nov 8 '15 at 8:03
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schematic

simulate this circuit – Schematic created using CircuitLab

In box 1, we're converting our current source into a voltage source and further converting it to its s-domain equivalent.

Then in box 2 we're taking the s-domain source together with the s-domain equivalent of the impedances of the given circuit. (see this)

Assuming that the capacitances had no voltage across them initially, (i.e. \$V_{0_{C_1}}=V_{0_{C_2}}=0\$, the s-domain circuit is equivalent to that in box 3 where \$V=-V_1\$ and \$R_{\text{equiv}}=\left[\left[\left(R_2+\frac{1}{sC_2}\right) \parallel\ R_L \right]\parallel \frac{1}{sC_1}\right]+R_1=R_1+\left[\frac{1}{R_2+sC_2}+\frac{1}{R_L}+\frac{1}{sC_1}\right]^{-1}\$

Hence $$V_1=-V=-\left(\frac{R_\text{equiv}}{R_s+R_\text{equiv}}\right)\cdot\frac{iR_s}{s}$$

The expression for \$V_1\$ will be a function of \$s\$; applying inverse laplace transform (using tables is easier) to the equation will give you the differential equation you require.

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