Question

Решить дифференциальное уравнение : y^2-9x^2+4xyy'=0

Answer 1

ЛОДУ 1-го порядка:
$y^2-9x^2+4xyy'=0\y=tx;y'=t'x+t\t^2x^2-9x^2+4x^2t(t'x+t)=0|:x^2\t^2-9+4tt'x+4t^2=0\9-5t^2=frac{4txdt}{dx}|*frac{dx}{x(9-5t^2)}\9-5t^2=0\5t^2=9\t^2=frac{9}{5}\t=^+_-frac{3}{sqrt5}\y=^+_-frac{3x}{sqrt5}\\y=frac{3x}{sqrt5}\frac{9x^2}{5}-9x^2+frac{36x^2}{5}=0\0=0\\y=-frac{3x}{sqrt5}\frac{9x^2}{5}-9x^2+frac{36x^2}{5}=0\0=0$
$frac{dx}{x}=4frac{tdt}{9-5t^2}\frac{dx}{x}=-frac{2}{5}frac{d(9-5t^2)}{9-5t^2}\intfrac{dx}{x}=-frac{2}{5}intfrac{d(9-5t^2)}{9-5t^2}\ln|x|=-frac{2}{5}ln|9-5t^2|+C\ln|x|+frac{2}{5}ln|frac{9x^2-5y^2}{x^2}|=C;y=^+_-frac{3sqrt5x}{5}$