Answer to Question #103574 in Differential Equations for Mahaveer yadav

Question

Solve : y^2 + x^2 (dy/dx)^2 -2xy (dy/dx) = 4 (dx/dy)^2

Answer 1

$y^2-2xyy'+x^2(y')^2-\frac{4}{(y')^2}=0\\ a=1, b=-2xy', c=x^2(y')^2-\frac{4}{(y')^2}\\ D=b^2-4ac=4x^2(y')^2-4(x^2(y')^2-\frac{4}{(y')^2})=\\ =\frac{16}{(y')^2}\\ 1) y=\frac{2xy'-\frac{4}{y'}}{2} \implies y=xy'-\frac{2}{y'} (*)$

this is the Clairaut's equation, make a substitution

$y'=p, dy=p dx$

then

$y=xp-\frac{2}{p}\\ dy=pdx +xdp+\frac{2}{p^2}dp\\ pdx=pdx +xdp+\frac{2}{p^2}dp\\ (x+\frac{2}{p^2})dp=0\\ a) dp=0 \implies p=c \implies y=xc-\frac{2}{c}$

is the general solution

$b) x+\frac{2}{p^2}=0$

let's say $x<0$ then

$p=\pm\sqrt\frac{2}{-x} \implies y'=\pm\sqrt\frac{2}{-x} \implies\\ y=\pm(-2\sqrt2 \sqrt{(-x)}) +c_1$

substitute y in (*)

$\pm(-2\sqrt2 \sqrt {( -x )})+c_1=\\ =x(\pm\sqrt\frac{2}{-x})-(\pm\frac{2}{\sqrt\frac{2}{-x}})\implies\\ c_1=0$

then $y=\pm(-2\sqrt2 \sqrt{(-x)})$ are solutions of the equation.

2) $y=\frac{2xy'+\frac{4}{y'}}{2} \implies y=xy'+\frac{2}{y'} (**)$

this is the Clairaut's equation, make a substitution

$y'=p, dy=p dx$

then

$y=xp+\frac{2}{p}\\ dy=pdx +xdp-\frac{2}{p^2}dp\\ pdx=pdx +xdp-\frac{2}{p^2}dp\\ (x-\frac{2}{p^2})dp=0\\ a) dp=0 \implies p=c \implies y=xc+\frac{2}{c}$

is the general solution

$b) x-\frac{2}{p^2}=0$

let's say $x>0$ then

$p=\pm\sqrt\frac{2}{x} \implies y'=\pm\sqrt\frac{2}{x} \implies\\ y=\pm2\sqrt2 \sqrt x +c_2$

substitute y in (**)

$\pm2\sqrt2 \sqrt x+c_2=\pm x\sqrt\frac{2}{x}+\frac{2}{\pm\sqrt\frac{2}{x}}\implies\\ c_2=0$

then $y=\pm2\sqrt2 \sqrt x$ are solutions of the equation.