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I have the following two equations, and I'm not sure I'm analyzing them properly. $ I(t)=CV'(t)\\ V(t)I(t) = P + R_{C}I^{2}(t)\\ \text{Substitute:}\\ CV(t)V'(t) = P + R_{C}C^{2}(V'(t))^{2}\\ $

If I'm right, this gives me a non-linear differential equation, which puts me well past my mathematical comfort zone.

Is this, in fact, a non-linear differential equation? If not, can anyone help alleviate my misunderstanding? If it IS non-linear, does anyone recognize this as an equation that has been/can be solved? Or am I out of luck as far as an exact solution goes?

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    @RahulNarain Absolutely. I need to formulate some thoughts first, however.2012-08-17

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Taking Fabian's suggestion, we switch the dependent and independent variables, giving $\frac{CV}{t'(V)} = P + \frac{R_CC^2}{t'(V)^2}$ because $V'(t) = \mathrm dV/\mathrm dt = (\mathrm dt/\mathrm dV)^{-1} = 1/t'(V)$. Since $t'(V)$ better not ever be zero, this is equivalent to $CVt'(V)=Pt'(V)^2+R_CC^2,$ which is a quadratic equation in $t'(V)$. Over Ed Gorcenski's remonstrations, we hit this with the quadratic-formula hammer, yielding $t'(V) = \frac{CV}{2P} \pm \sqrt{\left(\frac{CV}{2P}\right)^2-\frac{R_CC^2}P}.$ This can now be integrated as usual, using the fact that $\int\sqrt{x^2-a^2}\,\mathrm dx = \frac12\left(x\sqrt{x^2-a^2} - a^2\ln(x+\sqrt{x^2-a^2})\right).$ I haven't worked through by hand it myself, but Mathematica tells me that the integral simplifies to $t(V) = \frac{CV^2}{4P} \pm \frac{CV}{4P}\sqrt{V^2-4PR_C} \mp CR_C\log\big(V+\sqrt{V^2-4PR_C}\big) + \text{const},$ which looks to be the same as what Robert Israel got.

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Yes, it is definitely non-linear. Mathematica solves it, whereas WolframAlpha is not able to produce a formula. Here is the output of Mathematica 8.1, where I solved the equation $V(t)V'(t)=1+V'(t)^2$:

{{v[t] -> InverseFunction[ 1/8 (4 + 8 ArcSinh[Sqrt[-2 + #1]/2] - #1^2 - Sqrt[-2 + #1] #1 Sqrt[2 + #1]) &][-(t/2) + C[1]]}, {v[t] -> InverseFunction[ 1/8 (-4 + 8 ArcSinh[Sqrt[-2 + #1]/2] + #1^2 - Sqrt[-2 + #1] #1 Sqrt[2 + #1]) &][t/2 + C[1]]}}

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    Yes, but I posted my answer at 7pm and I was going to dinner, so I had no time to try :-)2012-08-17
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Maple 16's solution (switching to lowercase because I has a special meaning in Maple):

e1:= i(t) = c*diff(v(t),t); e2:= v(t)*i(t)=p+r[c]*i(t)^2; simplify(dsolve({e1,e2})); 

$\displaystyle [ \left\{ v \left( t \right) =r_{{c}}p \left( {\it LambertW} \left( -4\,r_{{c}}p{{\rm e}^{{\frac {-2\,t+r_{{c}}c+2\,{\it \_C1}}{r_{{c}}c}}}} \right) -1\\ \mbox{} \right) \left( \sqrt{-r_{{c}}p \left( {\it LambertW} \left( -4\,r_{{c}}p{{\rm e}^{{\frac {-2\,t+r_{{c}}c+2\,{\it \_C1}}{r_{{c}}c}}}}\\ \mbox{} \right) \right) ^{-1}} \right) ^{-1} \left( {\it LambertW} \left( -4\,r_{{c}}p{{\rm e}^{{\frac {-2\,t+r_{{c}}c+2\,{\it \_C1}}{r_{{c}}c}}}} \right) \right) ^{-1},v \left( t \right) =-r_{{c}}p \left( -1+{\it LambertW} \left( -1/4\,{{\rm e}^{{\frac {-2\,t+r_{{c}}c+2\,{\it \_C1}}{r_{{c}}c}}}}{r_{{c}}}^{-1}{p}^{-1} \right) \right) \sqrt{-{r_{{c}}}^{-1}{p}^{-1} \left( {\it LambertW} \left( -1/4\,{{\rm e}^{{\frac {-2\,t+r_{{c}}c+2\,{\it \_C1}}{r_{{c}}c}}}}{r_{{c}}}^{-1}{p}^{-1} \right) \right) ^{-1}}\\ \mbox{} \right\} , \left\{ i \left( t \right) =c{\frac {d}{dt}}v \left( t \right) \right\} ] $

It may actually be better to use the implicit form:

dsolve({e1,e2},implicit); 

$\displaystyle [ \left\{ t-1/4\,{\frac {c \left( v \left( t \right) \right) ^{2}}{p}}-1/4\,{\frac {cv \left( t \right) \sqrt{ \left( v \left( t \right) \right) ^{2}-4\,r_{{c}}p}}{p}}+r_{{c}}c\ln \left( v \left( t \right) + \sqrt{ \left( v \left( t \right) \right) ^{2}-4\,r_{{c}}p} \right) \\ \mbox{}-{\it \_C1}=0,t-1/4\,{\frac {c \left( v \left( t \right) \right) ^{2}}{p}}+1/4\,{\frac {cv \left( t \right) \sqrt{ \left( v \left( t \right) \right) ^{2}-4\,r_{{c}}p}}{p}}-r_{{c}}c\ln \left( v \left( t \right) + \sqrt{ \left( v \left( t \right) \right) ^{2}-4\,r_{{c}}p} \right) \\ \mbox{}-{\it \_C1}=0 \right\} , \left\{ i \left( t \right) =c{\frac {d}{dt}}v \left( t \right) \right\} ]$