differential equation,what is the general solution?

Question

please help ......i dont find the general solution.....itt is a linear differential equation

1. (y+1)dx + (4x-y)dy=0
2. udx + (1-3u)x du = 3udu
3. (y-x+xy cot x )dx +xdy = 0
4 .y' = csc x - ycot x

thanks for the help

Answer 1

waaaaaaaaaaaa.........its hard

1.[ (y+1)dx + (4x-y)dy=0 ]/(y+1)dy
dx/dy+4x/(y+1)-y/(y+1)=0
dx/dy+x(4/(y+1))=y/(y+1)
P(y)=4/(y+1) ; Q(y)=y/(y+1)
integrating factor = e^integrate.P(y)dy
e^4integrate dy/(y+1)
e^4ln(y+1)
=(y+1)^4
general soln
x(IF)= integrate Q(IF)dy
x(y+1)^4 = integrate y/(y+1)*(y+4)^4
= integrate y(y+1)^3
= integate( y^4+3y^4+3y^2+y)dy
x(y+1)^4 =y^5/5+y^3+y^2/2+c
ans: 10x(y+1)^4=8y^5+10y^3+5y^2+c

the rest try to solve it....just follow the format....^_^

Answer 2

I don't have reference materials handy but they all appear to be linear eqns.

The first step is to put them into the standard form of a linear first order eqn.

Answer 3

I made an A in Diff EQ, but that was in 1976. WOW, that has been a while. I have not clue but these look familiar.