Question

Дифференциальная уравнение ху'=1+у

Answer 1

$\begin{cases}R(x,y)=-y-1\\ S(x,y)=x\end{cases}\Rightarrow \left (\cfrac{\partial }{\partial y}R(x,y)=-1 \right )\neq \left ( 1=\cfrac{\partial }{\partial x}S(x,y) \right )$

$R(x,y)\mu (x)+\cfrac{\mathrm{d} y}{\mathrm{d} x}S(x,y)\mu (x)=0; \; \cfrac{\partial }{\partial y}(\mu (x)R(x,y))=\cfrac{\partial }{\partial x}(\mu (x)S(x,y))$

$-\mu (x)=x\cfrac{\mathrm{d} \mu (x)}{\mathrm{d} x}+\mu (x)\Leftrightarrow \cfrac{1}{\mu (x)}\cfrac{\mathrm{d} \mu (x)}{\mathrm{d} x}=-\cfrac{2}{x}\Leftrightarrow \ln \mu (x)=-2\ln x\Rightarrow \mu (x)=\cfrac{1}{x^2}$

$\begin{cases}P(x,y)=-\cfrac{y+1}{x^2}\\Q(x,y)=\cfrac{1}{x}\end{cases}\Rightarrow \cfrac{\partial }{\partial y}P(x,y)=-\cfrac{1}{x^2}=\cfrac{\partial }{\partial x}Q(x,y)$

$\begin{cases}\cfrac{\partial f(x,y)}{\partial x}=P(x,y)\\\cfrac{\partial f(x,y)}{\partial y}=Q(x,y)\end{cases}\Rightarrow f(x,y)=-\int \cfrac{y+1}{x^2}dx=\cfrac{y+1}{x}+g(y)$

$\cfrac{\partial f(x,y)}{\partial y}=\cfrac{\partial }{\partial y}\left ( \cfrac{y+1}{x}+g(y) \right )=\cfrac{1}{x}+\cfrac{\mathrm{d} f(y)}{\mathrm{d} y}$

$\cfrac{1}{x}+\cfrac{\mathrm{d} g(y)}{\mathrm{d} y}=\cfrac{1}{x}\Rightarrow \cfrac{\mathrm{d} g(y)}{\mathrm{d} y}=0\Rightarrow g(y)=0$

$f(x,y)=\cfrac{y+1}{x}\Rightarrow \cfrac{y+1}{x}=C\Rightarrow y=Cx-1$