Non-linear differential equation involving a function and its first and second derivatives I was wondering if anyone could get me started on solving the following:
$$(u')^2-u\cdot u''=0$$
I have tried letting $v=u'$, but I don't seem to make progress with such a substitution.
 A: Recalling the logarithmic derivative:
$$\frac{u''}{u'}=(\ln u')'=\frac{u'}{u}=(\ln u)'$$
Integration reduces it to a first order equation 
$\ln u'=\ln u +C$, or $u'=Bu$ ($B=e^C$). Hence the general solution is $u=Ae^{Bt}$. This also covers the special cases $u=0$ and $u'=0$ that were neglected when dividing.
The same can be done using the substitution $v=\frac{u'}{u}$. Then $v'=\frac{u''}{u}-v^2=\frac{u''}{u'}v-v^2$, so $v'=0$, and $\frac{u'}{u}=B$ again.
A: With
$(u')^2 - uu'', \tag 1$
I will develop a solution under the additional assumption
$u'u \ne 0; \tag 2$
from (1) we may write
$\dfrac{u'}{u} = \dfrac{u''}{u'}, \tag 3$
or
$(\ln u)' = (\ln u')'; \tag 4$
taking
$u(x_0) = u_0, \; u'(x_0) = u'_0, \tag 5$
we integrate (4) 'twixt $x_0$ and $x$:
$\ln u(x) - \ln u_0 = \displaystyle \int_{x_0}^x (\ln u(s))' \; ds = \int_{x_0}^x (\ln u'(s))' \; ds = \ln u'(x) - \ln u'_0; \tag 6$
that is,
$\ln \left (\dfrac{u(x)}{u_0} \right ) = \ln \left (\dfrac{u'(x)}{u'_0} \right ); \tag 7$
thus,
$\dfrac{u(x)}{u_0} =\dfrac{u'(x)}{u'_0}, \tag 8$
which again may be re-written
$\dfrac{u'(x)}{u(x)} = \dfrac{u'_0}{u_0}, \tag 9$
or
$(\ln u(x))' = \dfrac{u'_0}{u_0}; \tag{10}$
we integrate again
$\ln \left (\dfrac{u(x)}{u_0} \right ) = \ln u(x) - \ln u_0 = \displaystyle \int_{x_0}^x \dfrac{u'_0}{u_0} \; ds = \dfrac{u'_0}{u_0}  (x - x_0), \tag{11}$
whence
$\dfrac{u(x)}{u_0} = e^{(u'_0/u_0)(x - x_0)}, \tag{12}$
and finally
$u(x) = u_0e^{(u'_0/u_0)(x - x_0)}. \tag{13}$
This equation may be easily checked, for it yields
$u'(x) = u'_0 e^{(u'_0/u_0)(x - x_0)} \tag{14}$
and
$u''(x) = \dfrac{{u'}_0^2}{u_0}e^{(u'_0/u_0)(x - x_0)};\tag{15}$
it is clear from (13)-(15) that
$(u'(x))^2 = u''(x)u(x) = {u'}_0^2 e^{(2u'_0/u_0)(x - x_0)}. \tag{16}$
A: You can also observe that if you divide by $u'^2$ then it's a derivative
$$(u')^2-u\cdot u''=0 \implies \frac {(u')^2-u\cdot u''}{u'²}=0$$
$$ \left (\frac {u}{u'} \right )'=0$$
Integrate :
$$u=Ku' \implies \frac {u'}{u}=C \implies (\ln u)'=C$$
$$ \ln u = Cx+K \implies u(x)=Ke^{Cx}$$
