# 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?

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Apply the quadratic formula to get $V'(t) = \big(V(t)\pm\sqrt{V(t)^2-4PR_C}\big)\big/(2R_CC)$. It's still nonlinear and ugly, but now it's an ordinary differential equation. –  Rahul Narain Aug 16 '12 at 20:11
@RahulNarain I'm not entirely sure you can just do that. –  Arkamis Aug 16 '12 at 21:03
@Ed, I don't see why not. At any given $t$, $V(t)$ and $V'(t)$ are just real numbers, say $v$ and $w$ respectively. For $Cvw=P+R_CC^2w^2$ to hold, $w$ must be given by the quadratic formula. What else could it be? –  Rahul Narain Aug 16 '12 at 22:29
Because there exists a constraint that $V(t)^2 > 4PR_C$ for $P, R_C > 0$ that does not exist in the differential algebraic form. –  Arkamis Aug 16 '12 at 22:35
$I(t)$ is just a scaled version of $V'(t)$. If $V(\tau) < 4 PR_C$ for some $t = \tau$, then $I(t)$ would take a complex value. However, in the numerical solution, there are clearly values where $V(t) < 4PR_C$ and yet $I(t)$ is real-valued. That means that either the solution is wrong (which is possible), or that there is something else going on. However, the problem as originally formulated is a well-posed Index 1 DAE; ode15s should have no difficulty solving it. –  Arkamis Aug 16 '12 at 22:41

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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+1: I like your solution ;-) –  Fabian Aug 17 '12 at 9:37
@Fabian I liked your comment first! :) You should have posted an answer. –  Rahul Narain Aug 17 '12 at 12:47

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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The solution suggests that one should switch the dependent and independent variable and find an equation for $t(V)$ which then hopefully is separable. –  Fabian Aug 16 '12 at 17:40
Yes, but I posted my answer at 7pm and I was going to dinner, so I had no time to try :-) –  Siminore Aug 17 '12 at 8:20

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);


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

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