Solusi dr PD (1+y^2) sin x dx + 2y(1-cosx)dy=0 adalaaah..........

Materi Persamaan Diferensial <<<<<

[tex](1+y^2)\sin{x}\ \ dx+2y(1-\cos{x})\ \ dy=0\\2y(\cos{x}-1)dy=(1+y^2)\sin{x}dx\\\frac{2y}{1+y^2}dy=\frac{\sin{x}}{\cos{x}-1}dx\\\ln{\left(1+y^2\right)}=\ln{(\cos{x}-1)^{-1}}+C\\1+y^2=\exp{\left(\ln{(\cos{x}-1)^{-1}+C\right)}\\1+y^2=e^C\times \left(\cos{x}-1\right)\\y^2=K(\cos{x}-1)-1[/tex]