Предмет: Математика, автор: Аноним

Помогите решить
dy-cos(y-1)dx=0

Ответы

Автор ответа: NNNLLL54

$dy-cos(y-1)dx=0\\frac{dy}{cos(y-1)}=dx\\int frac{dy}{cos(y-1)}=int frac{cos(y-1)}{cos^2(y-1)}dy=int frac{coe(y-1)}{cos^2(y-1)}dy=int frac{cos(y-1)}{1-sin^2(y-1)}dy=\\=[t=sin(y-1),dt=cos(y-1)dy]=int frac{dt}{1-t^2}=frac{1}{2}ln|frac{1+t}{1-t}|+C=\\=frac{1}{2}ln|frac{1+sin(y-1)}{1+sin(y-1)}|+C\\\frac{1}{2}ln|frac{1+sin(y-1)}{1+sin(y-1)}|=x+C$

$frac{1+sin alpha }{1-sin alpha }=frac{1+cos(90^0- alpha )}{1+cos(90^0- alpha )}=frac{2cos^2(45^0-frac{ alpha }{2})}{2sin^2(45^0-frac{ alpha }{2})}=ctg^2(45^0-frac{alpha}{2})$

$frac{1}{2}ln|ctg^2(frac{pi}{4}-frac{y-1}{2})|=x+C\\ln|ctg(frac{pi}{4}-frac{y-1}{2})|=x+C$

$ctg(frac{pi}{4}- alpha )=tg(frac{pi}{2}-(frac{pi}{4}- alpha ))=tg(frac{pi}{4}+ alpha )\\Otvet:ln|tg(frac{pi}{4}+frac{y-1}{2})|=x+C$