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

We are asked to prove:

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 $t\to\pm\infty$.

Thanks in advance to anybody who will reply.

-Mario-

  • 5
    the right hand side is $(\sqrt{y})'$, so your equation is $y'' = (\sqrt{y})'$. Once you integrate both sides the resulting equation is manageable.2011-08-27

2 Answers 2

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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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    @user15123: I should also mention that $C=0$ ($v=\sqrt{u}$) admits the solution $y=\frac{1}{4}(t-B)^2$, which exists for all $t\in\mathbb{R}$.2011-08-29
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$\dfrac{d}{dx}\sqrt{y} = \dfrac{1}{2\sqrt{y}}\cdot \dfrac{dy}{dx}.$ (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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    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 $y'=p(y)$.2013-10-26