równania zupełne

[malineczka8888]

Znaleźć całkę ogólną równania zupełnego:
a)\((lny-2x)dx+( \frac{x}{y}-2y)dy=0\)

b)\((e^x+y)dx+(2+x+ye^y)dy=0\)

c)\(\frac{2x}{y^3}dx+ \frac{y^2-3x^2}{y^4}dy=0\)

d)\(x-y+(2y-x) \frac{dy}{dx}=0\)

[octahedron]

\(a)\,F(x,y)=\int \ln y-2x\,dx=x\ln y-x^2+D(y)
F_y=\frac{x}{y}+D'(y)=\frac{x}{y}-2y
D'(y)=-2y
D(y)=-y^2+C
F(x,y)=x\ln y-x^2-y^2=C\)

[octahedron]

\(b)\,F(x,y)=\int e^x+y\,dx=e^x+xy+D(y)
F_y=x+D'(y)=2+x+ye^y
D'(y)=2+ye^y
D=\int 2+ye^y\,dy=2y+ye^y-\int e^y\,dy=2y+(y-1)e^y+C
F(x,y)=e^x+xy+2y+(y-1)e^y=C\)

[octahedron]

\(c)\,F(x,y)=\int\frac{2x}{y^3}\,dx=\frac{x^2}{y^3}+D(y)
F_y=-\frac{3x^2}{y^4}+D'(y)=\frac{y^2-3x^2}{y^4}
D'(y)=\frac{1}{y^2}
D(y)=-\frac{1}{y}+C
F(x,y)=\frac{x^2}{y^3}-\frac{1}{y}=C\)

[octahedron]

\(d)\,F(x,y)=\int x-y\,dx=\frac{1}{2}x^2-xy+D(y)
F_y=-x+D'(y)=2y-x
D'(y)=2y
D(y)=y^2+C
F(x,y)=\frac{1}{2}x^2-xy+y^2=C\)