# Solving the 'easy' differential equation $(1 - \phi^2)\phi'' + \phi(\phi')^2 =0$.

I need to solve the following:

$$(1 - \phi^2)\phi'' + \phi(\phi')^2 =0.$$

Is there any standard method I can use?

• I don't think this is as easy as you think. – Ataraxia Jun 4 '13 at 22:14
• It may be harder :S I just saw two exponentials function in the Mathematica solution and i though it was easy :S – José D. Jun 4 '13 at 22:18

## 1 Answer

Just a lot of pattern matching and manipulation. Rewrite the equation as

$$\frac{\phi''}{\phi'} = -\frac{\phi \, \phi'}{1-\phi^2}$$

This can be written as

$$\frac{d}{dx} \log{\phi'} = \frac12 \frac{d}{dx} \log{(1-\phi^2)}$$

This may be integrated and subsequently exponentiated to produce

$$\phi' = A \left (1-\phi^2\right)^{1/2}$$

where $A$ is a constant of integration. We may then rewrite this equation in differential form as

$$\frac{d\phi}{\left (1-\phi^2\right)^{1/2}} = A \, dx$$

which integrates to

$$\arcsin{\phi} = A x + B$$

where $B$ is another constant of integration. The solution to the above equation is then

$$\phi(x) = \sin{(A x+B)}$$

You may verify that this is indeed the solution by plugging it back into the original equation.

• Mathematica's solution is $Cosh(C1 (t + C2))$ :S Is that right? – José D. Jun 4 '13 at 22:20
• @Trollkemada: You tell me. I plugged my solution into the original equation and the equation was satisfied. I don't see how your Mathematica solution works, unless your constants of integration are imaginary. – Ron Gordon Jun 4 '13 at 22:22
• You're right. I think both solutions are right. – José D. Jun 4 '13 at 22:23
• @RonGordon However, it's easy to verify that $\mathrm{cosh}(at+b)$ is indeed a solution of the original equation. When you integrate $u'/u$, don't forget you get $\log |u|$, not simply $\log u$. – Jean-Claude Arbaut May 16 '14 at 15:48