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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 $\lambda\not=0?$

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  • $\begingroup$ I edited your title to bring it into conformity with the body of your post, replacing $a$ by $\lambda$, and to properly $\LaTeX$ify it. Happy New Year! $\endgroup$ – Robert Lewis Jan 1 '18 at 17:37
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    $\begingroup$ Thanks Robert. Just fixed it, i had a small error in the title. $\endgroup$ – Y.L Jan 1 '18 at 18:35
  • $\begingroup$ My pleasure sir! Nice question, endorsed!!! $\endgroup$ – Robert Lewis Jan 1 '18 at 18:39
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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 $y(t)[y''(t)+2\lambda y'(t)]=(y'(t))^2$

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    $\begingroup$ Robert - Thank you for very useful answer ; $\endgroup$ – samjoe Jan 2 '18 at 15:12
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    $\begingroup$ Thanks a lot Robert, that is perfect! Thank you for the help. $\endgroup$ – Y.L Jan 3 '18 at 8:30
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    $\begingroup$ @Y.L If you find this answer useful then you can accept by clicking the tick mark on the left. cheers :) $\endgroup$ – samjoe Jan 3 '18 at 10:21
  • $\begingroup$ Thaks @samjoe i now did :) $\endgroup$ – Y.L Jan 4 '18 at 14:53
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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.

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  • $\begingroup$ I am also getting same answer as you, but what did you actually do in eqn 1? Thanks! $\endgroup$ – samjoe Jan 2 '18 at 2:56
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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 $\color{blue}{y = 0}$.

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  • $\begingroup$ What happens when $t = 0$, where $y(0) = 0$? $\endgroup$ – Robert Lewis Jan 1 '18 at 17:45
  • $\begingroup$ @robert I'm sorry does it cause a problem? I haven't calculated it yet $\endgroup$ – samjoe Jan 1 '18 at 17:47
  • $\begingroup$ I don't know for sure but $y(0) = 0$ is stipulated in the problem so I'm curious as to how you set up your equations near $t = 0$. But you got at least one upvote so someone must have thought it was OK. Cheers! $\endgroup$ – Robert Lewis Jan 1 '18 at 17:49
  • $\begingroup$ Right, you end up dividing by $0$ or having to deal with things like $0/0$. $\endgroup$ – Robert Lewis Jan 1 '18 at 17:50
  • $\begingroup$ @Robert how did you know this beforehand? I am getting log(0) .. is there a way around this? $\endgroup$ – samjoe Jan 1 '18 at 17:54

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