Question

Решить дифференциальное уравнение: (sqrtxy-x)dy+ydx=0

Answer 1

$(\sqrt{xy}-x)dy+y\, dx=0\, |:dy\\\\\sqrt{xy}-x+y\cdot \frac{dx}{dy}=0\\\\\frac{dx}{dy}=x'(y)\; \; \to \quad y\cdot x'-x=-\sqrt{xy}\, |:y\ne 0\\\\x'-\frac{x}{y}=-\frac{\sqrt{xy}}{y}\quad [\frac{\sqrt{xy}}{y}=\frac{\sqrt{x}\sqrt{y}}{y}=\frac{\sqrt{x}}{\sqrt{y}}]\\\\x'-\frac{x}{y}=-\sqrt{\frac{x}{y}}\\\\x=uv\; ,\; \; x'=u'v+uv'\\\\u'v+uv'-\frac{uv}{y}=-\sqrt{\frac{uv}{y}}\\\\u'v+u(v'- \frac{v}{y})=-\sqrt{\frac{uv}{y}}\\\\a)\; \; \frac{dv}{dy}-\frac{v}{y}=0\; ,\; \; \int \frac{dv}{v}=\int \frac{dy}{y}\; ,\; \; ln|v|=ln|y|\; ,\; \; v=y\\\\b)\; \; u'\cdot y=-\sqrt{\frac{uy}{y}}$

$\frac{du}{dy}\cdot y=-\sqrt{u}\\\\\int \frac{du}{\sqrt{u}}=-\int \frac{dy}{y}\\\\2\sqrt{u}=-ln|y|+lnC\\\\\sqrt{u}=\frac{1}{2}\cdot ln|\frac{C}{y}|\\\\u=\frac{1}{4}ln^2|\frac{C}{y}|$

$c)\; \; x=uv\\\\x=\frac{y}{4}\cdot ln^2| \frac{C}{y}|$