Is this a non-linear differential equation? And is there a solution? 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?
 A: 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.
A: 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]]}}
A: 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);

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