Show that  y^2 = (4x)(a-x) = 4ax-4x^2  is a solution to the DE?  2xy dy/dx = y^2 - 4x^2

(portions of this question have been edited or deleted!)

Oct 10, 2017

As the quoted Differential Equation is malformed, let us work back from the quoted solution to form the correct DE:

Correction of the Question:

If a solution is:

${y}^{2} = \left(4 x\right) \left(a - x\right) = 4 a x - 4 {x}^{2}$

is a solution, then implicit differentiation gives:

$\setminus \setminus \setminus \setminus \setminus 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 4 a - 8 x$

Multiplying by $x$ gives us:

$\setminus \setminus \setminus \setminus \setminus 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 4 a x - 8 {x}^{2}$
$\therefore 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = 4 a x - 4 {x}^{2} - 4 {x}^{2}$
$\therefore 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} - 4 {x}^{2}$

Therefore the question should (presumably) be to solve the DE

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} - 4 {x}^{2}$

And $x = \frac{a}{2} \implies {y}^{2} = \left(4 \frac{a}{2}\right) \left(a - \frac{a}{2}\right) = {a}^{2} \implies y = a$

Solution:

We have:

$2 x y \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} - 4 {x}^{2}$ ..... [A]

As suggested, let us perform the substitution:

$y = v x \iff v = \frac{y}{x}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = v \left(\frac{d}{\mathrm{dx}} x\right) + \left(\frac{d}{\mathrm{dx}} v\right) x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

Substituting into the (corrected) DE [A] we have:

$2 x \left(v x\right) \left(v + x \frac{\mathrm{dv}}{\mathrm{dx}}\right) = {\left(v x\right)}^{2} - 4 {x}^{2}$

$\therefore 2 {v}^{2} {x}^{2} + 2 v {x}^{3} \frac{\mathrm{dv}}{\mathrm{dx}} = {v}^{2} {x}^{2} - 4 {x}^{2}$

$\therefore 2 {v}^{2} + 2 v x \frac{\mathrm{dv}}{\mathrm{dx}} = {v}^{2} - 4$

$\therefore 2 v x \frac{\mathrm{dv}}{\mathrm{dx}} = - \left({v}^{2} + 4\right)$

$\therefore \frac{2 v}{{v}^{2} + 4} \frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{x}$

This is now a First Order separable DE, so we can "separate the variables" to get:

$\int \setminus \frac{2 v}{{v}^{2} + 4} \setminus \mathrm{dv} = - \setminus \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

And now we can integrate, to get:

$\therefore \ln \left({v}^{2} + 4\right) = - \ln x + C$

Applying the initial conditions:

$y = a$ when $x = \frac{a}{2}$
$y = v x \implies a = \frac{v a}{2}$
$\therefore v = 2$

So we have modified initial conditions $v = 2$ when $x = \frac{a}{2}$:

$\ln \left(4 + 4\right) = - \ln \left(\frac{a}{2}\right) + C \implies C = \ln \left(\frac{a}{2}\right) + \ln 8 = \ln 4 a$

So our particular solution is:

$\ln \left({v}^{2} + 4\right) = - \ln x + \ln \left(4 a\right)$

$\therefore \ln \left({v}^{2} + 4\right) = \ln \left(\frac{4 a}{x}\right)$

$\therefore {v}^{2} + 4 = \frac{4 a}{x}$

Restoring the substitution:

${\left(\frac{y}{x}\right)}^{2} + 4 = \frac{4 a}{x}$

$\therefore {y}^{2} / {x}^{2} + 4 = \frac{4 a}{x}$

$\therefore {y}^{2} + 4 {x}^{2} = 4 a x$

$\therefore {y}^{2} = 4 a x - 4 {x}^{2}$

$\therefore {y}^{2} = 4 x \left(a - x\right) \setminus \setminus \setminus$ QED