Second order nonlinear differential equation $y'' = C/y^{1/2}$

Question

$y'' = \frac{C}{\sqrt{y}}$

How can this differential equation be solved? I would love to add details about an attempt, but I have no idea where to even start.

Answer 1

Rewrite as $y'' = Cy^{-1/2}$

$y'' - Cy^{-1/2} = 0$

For variation of parameters assume the solution is in the same family of functions as the given derivative form. Since we are given a very simple power of x, try a simple power of x first.

Suppose $y = x^n$. Then $y'' = (n)(n-1)x^{n-2}$ and $Cy^{-1/2} = Cx^{-n/2}$

So $C = (n)(n-1)$ and $ n - 2 = -n/2$

$2n - 4 = -n$

$3n = 4$

$n = 4/3$

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Check: $n-2 = 4/3 - 6/3 = -2/3$ and $-n/2 = (-1/2)(4/3)= -2/3 = n-2$ as required.

Any multiple $K_1 x^{4/3}$ will also satisfy the equation.

$C = n(n-1)K_1 = (4/3)(1/3)K_1 = 4/9K_1$

Part of homogeneous solution $y_h = K_1 x^{4/3}$ and $y'' = 4/9K_1 x^{-2/3} = 4/9 y^{-1/2}$

Going a little further, trying to get a more general case, try a linear expression instead of just x. Suppose $y = (ax + b)^n$

Then $y'' = a^2(n)(n-1) (ax + b)^{n-2}$

So to make $y'' = Cy^{-1/2} = C (ax + b)^{-n/2}$ again we have the solution $n-2 = -n/2, n = 4/3$with $C = 4/9a^2$

$y_h = (ax + b)^{4/3}$

Here the two undetermined constants are a and b, as required for a second-order equation.

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Now we need a particular solution

I'm not getting what I need here and am out of time -- going to give you this partial solution as at least a place to start.

Some options for a particular solution might be sin^{4/3}x or $e^{kix}$. Something to work on,

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General solution $ y = y_p + y_h$

Answer 2

$$y'' = \frac{C}{\sqrt{y}}$$ This second order ODE is of autonomous kind. The usual method to reduce the order is shown below. The first order ODE obtained is separable.

The integration is easier considering the inverse function $x(y)$ instead of $y(x)$ which leads to : $$x+c_2=\pm\frac{1}{C^3}(C\sqrt{y}-c_1)\sqrt{2C\sqrt{y}+c_1}$$ Inversing $x(y)$ for $y(x)$ leads to a three degrees polynomial equation. Of course, analytically solvable, but tiresome.