Question

Решить однородное диф.уравнение

(x²-2xy)dy-(xy-y²)dx=0

И y'=y/x+e^(y/x)

Answer 1

Ответ:

Однородные дифференциальные уравнения 1 порядка .

$\displaystyle \bf1)\ \ (x^2-2xy)\, dy-(xy-y^2)\, dx=0\ \ \ \Rightarrow \\\\\frac{dy}{dx}=\frac{xy-y^2}{x^2-2xy}\ \ ,\ \ \ y'=\frac{\dfrac{y}{x}-\dfrac{y^2}{x^2}}{1-\dfrac{2y}{x}}$  

Замена :   $\displaystyle \bf u=\frac{y}{x}\ \ \to \ \ \ ,\ \ y=ux\ ,\ \ y'=u'x+u$    

$\displaystyle \bf u'x+u=\frac{u-u^2}{1-2u}\ \ \ ,\ \ u'x=\frac{u-u^2}{1-2u}-u\ \ ,\ \ \ u'x=\frac{u-u^2-u+2u^2}{1-2u}\ ,\\\\\\u'x=\frac{u^2}{1-2u}\ \ ,\ \ \frac{du}{dx}=\frac{u^2}{(1-2u)\cdot x}\ \ ,\\\\\\\int \frac{(1-2u)\, du}{u^2}=\int \frac{dx}{x}\ \ \ \Rightarrow \ \ \ \int \frac{du}{u^2}-2\int \frac{du}{u}=\int \frac{dx}{x}\ \ ,\\\\\\-\frac{1}{u}-2\cdot ln|u|=ln|x|+C\ \ \Rightarrow \ \ \ -\frac{x}{y}=2\cdot ln\Big|\, \frac{y}{x}\, \Big|=ln|x|+C$  

$\displaystyle \bf 2)\ \ y'=\frac{y}{x}+e^{\frac{y}{x}}\ ,\\\\Zamena:\ u=\frac{y}{x}\ \ ,\ \ y=ux\ \ ,\ \ y'=u'x+u\\\\u'x+u=u+e^{u}\ \ \Righyarrow \ \ \ u'x=e^{u}\ \ ,\ \ \frac{du}{dx}= \frac{e^{u}}{x}\ \ ,\\\\\int e^{-u}\, du=\int \frac{dx}{x}\ \ \ ,\\\\-e^{-u}=ln|x|+C\ \ \ \Rightarrow \ \ \ e^{\frac{y}{x}}=-ln|x|-C$