$ff'' - f '^2 = 1$ with $f$ a twice differentiable function, defined on $\mathbb R$.

Question

$$ff''- f'^2=1$$ From that I showed that $f$ is different from zero on $\mathbb{R}$ and that $\frac{f''}{f}$ is constant on $\mathbb{R}$, so I have: $$f'^2-\frac{f''}{f}\cdot f^2 = 1$$ and I don't know what to do next.

Answer 1

One trick you can always try with autonomous second order equations is to assume that some reduction to a first order equation exists, $f'=u(f)$. Then $f''=u'(f)u(f)$ and the equation reads $$ u'uf-u^2=1 $$ where it is more visible that the equation is separable, $$ \frac{u'u}{1+u^2}=\frac1f\implies 1+u^2=k^2f^2. $$

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With this as guide one can reverse engineer the transformations to get the same without the additional function, $$ \frac{2f''f'}{1+f'^2}=\frac{2f'}f $$ which again has easily integrable expressions on both sides.

Answer 2

Write $f''/f=k^2$ so $f=f_0\cosh kx+k^{-1}f_0'\sinh kx$ and $$1=k^2f^2-f'^2=(k^2f_0^2-f_0'^2)(\cosh^2 kx-\sinh^2 kx)=k^2f_0^2-f_0'^2.$$This gives the parameterisation$$f_0=k^{-1}\cosh\theta,\,f_0'=\sin\theta\implies f=k^{-1}\cosh (kx+\theta).$$

Answer 3

$$ff''- f'^2=1$$ Susbstitute $$p=f' \implies f''=\frac {df'}{dx}=\frac {df'}{df}\frac {df}{dx}=p'p$$ $$fpp'-p^2=1$$ It's separable $$\implies fpp'=1+p^2 \implies \int \frac {pdp}{1+p^2}=\int \frac {df}f$$

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Here is another approach $$ff''- f'^2=1$$ $$\implies (\frac {f'}f)'=\frac 1 {f^2}$$ $$d(\frac {f'}f)=\frac {dx} {f^2}$$ $$\frac {f'}fd(\frac {f'}f)=\frac {f'}f\frac {dx} {f^2}$$ $$\frac {f'}fd(\frac {f'}f)=\frac {df} {f^3}$$ $$\int \frac {f'}fd(\frac {f'}f)=\int \frac {df} {f^3}$$

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You can also integrate it this way $$fpp'-p^2=1$$

$$(p^2)'\frac f 2-p^2=1$$ Substitute $v=p^2$ $$v'-\frac 2fv=\frac 2f$$ This last DE is linear of first order