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1) 3 + cos(y)*y' - 5*y' = 0;
3 = y'(5 - cos(y));
y' = 3/(5 - cos(y));
2)y'' = -(3*(sin(y)*y')/(5-cos(y))^2 = (-9*sin(y))/(5 - cos(y))^3
4а - 2(8-2а)= 4а -16 +4а=8а-16
6(3а-b)-(а-3б)=18а-6б-а+3б=17а-2б
2(8а-7б)+4(а-2б)=16а-14б+4а-6б=20а-10б