differential equation $y(t)[y''(t)+2\lambda y'(t)]=(y'(t))^2$ Can anyone please help me solve the following differential equation
$$y(t)[y''(t)+2\lambda y'(t)]=(y'(t))^2$$
with $y(0)=0$.
If $\lambda=0$ then clearly $y(t)=Ge^{\alpha t}$. 
But what if $\lambda\not=0?$
 A: Provided the solution is not identical to the $y=0$ solution divide by $yy'$ at those solution segments where this is not zero to get
$$
\frac{y''}{y'}+2λ=\frac{y'}{y}\implies \ln|y'|+2λt=\ln|y|+c
$$
or
$$
y'=Ce^{-2λt}y.
$$
The next integration gives
$$
\ln|y|=-\frac{C}{2λ}e^{-2λt}+d
$$
or, collapsing constants,
$$
y(t)=De^{Ce^{-2λt}}
$$
However, with $y(0)=De^C$ the condition $y(0)=0$ is only possible for $D=0$, that is, the zero solution. Any other solution can not reach from a non-zero to the zero value in finite time.
A: First of all, I wish to use the $\dot{}$ notation for $d/dt$, since I think it will make things a little easier to read, since we have to deal with expressions like $(y'(t))^2 = \dot y^2$ etc.
Now assume for the moment there is some $\tau \in \Bbb R$ with
$y(\tau) \ne 0 \ne \dot y(\tau); \tag 1$
then if we provisionally grant that $y(t)$ and $\dot y(t)$ are continuous near such a $\tau$, so there is some interval $B(\tau, \epsilon) = (\tau - \epsilon, \tau + \epsilon)$ in which
$y(t) \ne 0 \ne \dot y(t), \; t \in B(\tau, \epsilon), \tag 2$
then we can analyze the system
$y(t)[\ddot y(t) + 2\lambda \dot y(t)] = \dot y^2(t) \tag 3$
by dividing through by $y(t) \dot y(t)$ to obtain
$\dfrac{\ddot y(t)}{\dot y(t)} + 2\lambda = \dfrac{\dot y(t)}{y(t)}; \tag 4$
we next observe that
$\dfrac{d(\ln \dot y(t))}{dt} = \dfrac{\ddot y(t)}{\dot y(t)}, \tag 5$
$\dfrac{d(\ln y(t))}{dt} = \dfrac{\dot y(t)}{ y(t)}; \tag 6$
substituting (5) and (6) into (4) yields, after a li'l tad o' algebra,
$\dfrac{d(\ln \dot y(t))}{dt} - \dfrac{d(\ln y(t))}{dt} = -2\lambda, \tag 7$
or
$\dfrac{d}{dt}(\ln \dot y(t) - \ln y(t)) = -2\lambda, \tag 8$
or
$\dfrac{d}{dt}(\ln \dfrac{\dot y(t)}{y(t)}) = -2\lambda; \tag 9$
we may integrate (9) 'twixt $\tau$ and $t$:
$\ln \dfrac{\dot y(t)}{y(t)} - \ln \dfrac{\dot y(\tau)}{y(\tau)} = \displaystyle \int_\tau^t \dfrac{d}{ds}(\ln \dfrac{\dot y(s)}{y(s)}) \; ds = -2\lambda(t - \tau), \tag{10}$
whence
$\ln \dfrac{\dot y(t)}{y(t)} = \ln \dfrac{\dot y(\tau)}{y(\tau)} - 2\lambda(t - \tau); \tag{11}$
we exponentiate each side:
$\dfrac{\dot y(t)}{y(t)} = e^{ \ln \dfrac{\dot y(\tau)}{y(\tau)}} e^{ - 2\lambda(t - \tau)} = \dfrac{\dot y(\tau)}{y(\tau)}  e^{ - 2\lambda(t - \tau)}; \tag{13}$
next, via (6),
$\dfrac{d(\ln y(t))}{dt} = \dfrac{\dot y(\tau)}{y(\tau)}  e^{ - 2\lambda(t - \tau)}; \tag{14}$
integrating once more between $\tau$ and $t$:
$\ln \dfrac{y(t)}{y(\tau)} = \ln y(t) - \ln y(\tau) = \displaystyle \int_\tau^t \dfrac{d(\ln y(s))}{ds} \; ds = -\dfrac{\dot y(\tau)}{2\lambda y(\tau)}(e^{ - 2\lambda(t - \tau)} - 1), \tag{15}$
whence
$\ln y(t) = \ln y(\tau) -\dfrac{\dot y(\tau)}{2\lambda y(\tau)}(e^{ - 2\lambda(t - \tau)} - 1); \tag{16}$
if we exponentiate one more time we obtain
$y(t) = e^{\ln y(\tau)} \exp(-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}(e^{ - 2\lambda(t - \tau)} - 1))$
$= y(\tau) \exp(-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}(e^{ - 2\lambda(t - \tau)} - 1)), \tag{17}$
which may also be written in the form
$y(t) = y(\tau) \exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)}) \exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)}), \tag{18}$
which is as far as we will take it here.  We check (18):
$\dot y(t) =  \dot y(\tau) \exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)}) \exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)})e^{ - 2\lambda(t - \tau)}, \tag{19}$
$\ddot y(t)$
$= \dot y(\tau) \exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)})\exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)}) \dfrac{\dot y(\tau)}{y(\tau)} e^{ - 4\lambda(t - \tau)}$
$- 2\lambda \dot y(\tau) \exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)}) \exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)})e^{ - 2\lambda(t - \tau)}; \tag{20}$
it is thus easy to see that
$\ddot y(t) + 2\lambda \dot y(t) = \dot y(\tau) \exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)})\exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)}) \dfrac{\dot y(\tau)}{y(\tau)} e^{ - 4\lambda(t - \tau)}$
$= \dfrac{\dot y^2(\tau)}{y(\tau)}\exp(\dfrac{\dot y(\tau)}{2\lambda y(\tau)})\exp (-\dfrac{\dot y(\tau)}{2\lambda y(\tau)}e^{ - 2\lambda(t - \tau)}) e^{ - 4\lambda(t - \tau)}; \tag{21}$
thus,
$y(t)[\ddot y(t) + 2\lambda y(t)] = \dot y^2(\tau) \exp(\dfrac{\dot y(\tau)}{\lambda y(\tau)})\exp (-\dfrac{\dot y(\tau)}{\lambda y(\tau)}e^{ - 2\lambda(t - \tau)}) e^{ - 4\lambda(t - \tau)}; \tag{22}$
it is now relativey easy to see, by comparison of (22) with (19), that $y(t)$ satisfies (3).
Though we assumed $\dot y(\tau) \ne 0$ for the purposes of deriving (18), we see from these last few equations that the solution we have derived is valid for all $\dot y(\tau)$.  In the event that $y(\tau) = 0$, (18) is no longer valid as it stands, but we see from (3) that $\dot y(\tau) =  0$ at such $\tau$, and thus that $y(t) = 0$ for all $t \in B(\tau, \epsilon)$ is a legitimate solution.  Furthermore, if $y(\tau) \ne 0$, the exponential nature of (18) shows that $y(t)$ can never vanish.  
To summarize, (18) presents the generic solution for $y(\tau) \ne 0$; if $y(\tau) = 0$, then $y(t) = 0$ is a solution, and for all $t$; since any solution with $y(\tau') \ne 0$ for some $\tau'$ can never reach $0$, the zero solution is unique.  
differential equation $y(t)[y''(t)+2\lambda y'(t)]=(y'(t))^2$
A: First write $y'' = y'\frac{dy'}{dy}$ using chain rule. So we get the following on simplification
$$ yy'\frac{dy'}{dy} + 2\lambda y' y = (y')^2 \\
\frac{dy'}{dy}-\frac{y'}{y}= -2\lambda$$
Now substitute $z=y'/y$ or just observe it's a linear differential equation.
So in this edit, I provide the further solution. The integration factor here is $e^{-\ln(y)} = 1/y$, so we get
$$\frac{y'}{y} = \int\frac{-2\lambda}{y}dy = -2\lambda \ln(Cy)$$
Therefore we have the following equation:
$$\begin{align}
\int\frac{dy}{ y\ln(Cy) } &= \int -2\lambda dt \\
 \ln(\ln(Cy)) + \ln(D) &= -2\lambda t \\
 \ln(D\ln(Cy)) &= -2\lambda t \\
\implies y &= C_1\exp(C_2\exp(-2\lambda t))
\end{align}$$
Where I have converted $D = \frac{1}{C_2}$ and $C=\frac{1}{C_1}$.
From here we note that $y(0) = 0$ is possible only when $C_1 = 0$. So only solution is $\color{blue}{y = 0}$.
