Question

Помогите решить уравнение срочно плиииз
20б

Answer 1

Ответ:

$y\, dx=(x-\sqrt{xy})\, dy\\\\\dfrac{dx}{dy}=\dfrac{x-\sqrt{xy}}{y}\ \ ,\ \ \ \displaystyle x'-\frac{x}{y}=\frac{\sqrt{x}}{\sqrt{y}}\ \ ,\\\\\\x=uv\ \ ,\ \ x'=u'v+uv'\ \ ,\ \ u=u(y)\ ,\ v=v(y)\ ,\\\\\\u'v+uv'-\frac{uv}{y}=\frac{\sqrt{uv}}{\sqrt{y}}\ \ ,\ \ \ u'v+u\cdot \Big(v'-\frac{v}{y}\Big)=\frac{\sqrt{x}}{\sqrt{y}}\ \ ,\\\\\\a)\ \ v'-\frac{v}{y}=0\ \ ,\ \ \frac{dv}{dy}=\frac{v}{y}\ \ ,\ \ \ \int \frac{dv}{v}=\int \frac{dy}{y}\ \ ,\ \ \ ln|v|=ln|y|\ \ ,\ \underline {\ v=y\ }$

$b)\ \ u'v=\dfrac{\sqrt{uv}}{\sqrt{y}}\ \ ,\ \ \displaystyle \frac{du}{dy}\cdot y=\frac{\sqrt{u}\cdot \sqrt{y}}{\sqrt{y}}\ \ ,\ \ \int \frac{du}{\sqrt{u}}=\int \frac{dy}{y}\ \ ,\\\\\\2\sqrt{u}=ln|y|+lnC\ \ \ ,\ \ 2\sqrt{u}=ln|Cy|\ \ ,\ \ \ \sqrt{u}=ln\sqrt{|Cy|}\ \ ,\ \ \underline {\ u=ln^2\sqrt{|Cy|}\ }\\\\\\c)\ \ x=uv\ \ ,\ \ \ \ \underline{\ x=y\cdot ln^2\sqrt{|Cy|}\ }$