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I have two equations and I'm trying to solve $V_2$. The equations are: $\frac{V_1}{sL}+\frac{V_1-V_2}{R}=I_s,\\ \frac{V_2-V_1}{R}+\frac{V_2}{1/sC} = 0$

If I solve the second equation with respect to $V_1$ I get: $V_1=V_2 (sRC+1)$

It all goes down hill when I sub $V_1$ into first equation. I end up getting $\frac{I_s}{\frac{1}{sL}+\frac{sC}{R}} $

However the correct answer is: $\frac{s I_s}{C[s^2+\frac{R}{L}s +\frac{1}{LC}]}$

Can someone help me walk through this step by step to figure out why I am getting the wrong answer.

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    I find maxima (wxmaxima) useful for symbolic calculations. A bit cumbersome, but free.2012-10-24

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When I plug $V_1=V_2(sRC+1)$ and $V_1-V_2=V_2sRC$ into $\frac{V_1}{sL}+\frac{V_1-V_2}{R}=I_s$ I get $ \begin{align} V_2 \left(\frac{sRC+1}{sL}+sC\right)&=I_s\\ V_2 \frac{sRC+1+s^2LC}{sL} &= I_s \end{align} $ which means $V_2= \frac{I_ssL}{sRC+1+s^2LC}.$ For some reason, in the answer from your book they continued to get $V_2= \frac{I_ssL}{sRC+1+s^2LC} = \frac{I_ssL}{LC \left(\frac{sR}L + \frac1{LC} + s^2\right)} = \frac{I_ss}{C \left(\frac{sR}L + \frac1{LC} + s^2\right)}.$

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    Yes this is algebra for an output voltage using the laplace transform.2012-10-24