# Differential equation $y''=\dfrac{y'}{2\sqrt y}$

I'm struggling in solving this problem. We are given the differential equation $\displaystyle y''=\dfrac{y'}{2\sqrt{y}}$.

a) Any nonconstant solution is strictly monotone.

b) Let's consider the Cauchy Problem

$$\begin{cases}y''&=&\dfrac{y'}{2\sqrt y}\\[6pt] y(0)&=&u\\[6pt] y'(0)&=&v.\end{cases}$$

Find all points $(u,v)\in\mathbb R_+\times \mathbb R$ such that the maximal solution is nonconstant and defined on the whole real line. For such points $(u,v)$ compute the limits of the solution as $t\to\pm\infty$.

-Mario-

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What did you try? – Davide Giraudo Aug 27 '11 at 15:00
the right hand side is $(\sqrt{y})'$, so your equation is $y'' = (\sqrt{y})'$. Once you integrate both sides the resulting equation is manageable. – Zarrax Aug 27 '11 at 15:07

Try integrating: $y'=C+\sqrt{y}$. This should be easier to solve.

Additionally, note that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\log(|y'|)=\frac{y''}{y'}=\frac{1}{2\sqrt{y}}\ge0$. This means that $|y'|$ is monotonically increasing. So if $y'<0$ at some point, $y'$ decreases monotonically, and if $y'>0$ at some point, $y'$ increases monotonically.

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Uhm... i think that from $\frac{\mathrm d}{\mathrm d t}\log(|y'|)=\frac{\mbox{sgn}(y')}{2\sqrt y}$ we can't conclude what you did. I mean.. you seemed to forget the sign term but considering it we can't conclude about the monotonicity of the absolute value of $y'$. Am i wrong? – uforoboa Aug 28 '11 at 12:04
@user15123: I believe that if you take the derivative of $\log(|x|)$, you will see that it is $1/x$, whether $x>0$ or $x<0$. – robjohn Aug 28 '11 at 18:23
ops.... you are right.. however i've tried to work about the equation $y'=\sqrt y+C$ but I didn't come up with anything interesting to solve the problem. Can you develop your idea please? – uforoboa Aug 28 '11 at 18:59
@user15123: it is possible to solve for $y$ in terms of $t$ using the Lambert $W$-function. However, we only need to solve $y'=C+\sqrt{y}$ to get $t=2\sqrt{y}-2C\log(|C+\sqrt{y}|)+B$. If $t$ is going to range over all $\mathbb{R}$, we need $C<0$ so that $|C+\sqrt{y}|$ can go to $0$ where $t$ will go to $-\infty$. Note that $C=v-\sqrt{u}$. Therefore, $C<0$ iff $v<\sqrt{u}$. As $t\to+\infty$, $y\to+\infty$ (like $t^2/4)$. As $t\to-\infty$, $y\searrow C^2$. – robjohn Aug 29 '11 at 12:55
thanks a lot.. really appreciated your patience and kindness. Best wishes for everything robjohn. Thanks again and goodbye :D – uforoboa Aug 29 '11 at 13:49

$\dfrac{d}{dx}\sqrt{y} = \dfrac{1}{2\sqrt{y}}\cdot \dfrac{dy}{dx}.$ (This is an application of the chain rule.)

So $y'' = \dfrac{y'}{2\sqrt{y}}$ is the same as $\dfrac{d}{dx} \dfrac{d}{dx} y = \dfrac{d}{dx}\sqrt{y}$.

It follows that $\dfrac{d}{dx}y = \sqrt{y} + C$.

So $\dfrac{dy}{\sqrt{y} + C} = dx$.

$\displaystyle \int \dfrac{dy}{\sqrt{y} + C} = \int dx$.

If $u=\sqrt{y}$, then $u^2 = y$, so $2u\;du = dy$, and we get $$\int \frac{2u\;du}{u + C} = \int dx.$$ $$\int 2 - \frac{2C}{u+C} du = x + B.$$ etc.

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I'm sorry i can't go on with your calculations. Besides, i'm not seeing any reasonable way to solve the problem starting with your idea. Can you show me more please? Thanks. – uforoboa Aug 28 '11 at 13:13
Any DE of the form $F(y, y', y'')=0$ can be reduced to a first-order DE $F(y, p, p')=0$ by means of the substitution $y'=p(y)$. – daulomb Oct 26 '13 at 17:06