jak rozwiązać opis: Równanie różniczkowe

	dy		1+x sin y
−x2		=
	dx		1+cos y

Mariusz: −x²(1+cos(y))dy=(1+x sin(y))dx (1+x sin(y))dx+x²(1+cos(y))dy=0

	dy
(1+x sin(y))+x2(1+cos(y))		=0
	dx

cos(y)=(1−sin(y))u cos²(y)=(1−sin(y))²u² (1−sin²(y))=(1−sin(y))²u² (1−sin(y))(1+sin(y))=(1−sin(y))²u² (1+sin(y))=(1−sin(y))u² 1+sin(y)=u² − u² sin(y) sin(y)(1+u²)=u²−1

	u2−1
sin(y)=
	u2+1

	u2−1
cos(y)=(1−		)u
	u2+1

	2u
cos(y)=
	u2+1

	dy		2u(u2+1)−2u(u2−1)	du
cos(y)		=
	dx		(u2+1)2	dx

	dy		2u(u2+1−u2+1)	du
cos(y)		=
	dx		(u2+1)2	dx

	dy		2u	2	du
cos(y)		=
	dx		u2+1	u2+1	dx

2u	dy		2u	2	du
		=
u2+1	dx		u2+1	u2+1	dx

dy		2	du
	=
dx		u2+1	dx

	u2−1		2u		2	du
(1+x		)+x2(1+		)			=0
	u2+1		u2+1		u2+1	dx

	2u		du
(u2+1+x (u2−1))+2x2(1+		)		=0
	u2+1		dx

	(1+u)2		du
(u2+1+x (u2−1))+2x2(		)		=0
	u2+1		dx

	du
((u2+1)2+x (u4−1)) +2x2(1+u)2		=0
	dx

To podstawienie pozwoliło pozbyć się funkcji trygonometrycznych

Mariusz:

	dy		1+xsin(y)
−x2		=
	dx		1+cos(y)

	dy
−x2(1+cos(y))		=1+xsin(y)
	dx

−x²(1+cos(y))dy=(1+xsin(y))dx (1+xsin(y))dx+x²(1+cos(y))dy 1+xsin(y)=u 1+cos(y)=v xsin(y)=u−1 cos(y)=v−1

	u−1
x=
	sin(y)

y=arccos(v−1)

	u−1
x=
	√1−(v−1)2

y=arccos(v−1)

	u−1
x=
	√2v−v2

y=arccos(v−1)

u		u(u−1)(v−1)		(u−1)2v	1
	du+(		−			)dv=0
√2v−v2		(2v−v2)√2v−v2		2v−v2	√2v−v2

u		(u−1)(v−u)
	du+		dv=0
√2v−v2		(2v−v2)√2v−v2

u(2v−v²)du+(u−1)(v−u)dv=0 −uv(v−2)du+(u−1)(v−u)dv=0