Non-linear differential equation involving a function and its first and second derivatives

Question

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.

Answer 1

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

$(u'(x))^2 = u''(x)u(x) = {u'}_0^2 e^{(2u'_0/u_0)(x - x_0)}. \tag{16}$

Answer 2

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.

Answer 3

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 : $$u=Ku' \implies \frac {u'}{u}=C \implies (\ln u)'=C$$ $$ \ln u = Cx+K \implies u(x)=Ke^{Cx}$$