C is an arbitrary constant, from the choices which would be a solution to the differential equation xy'+4y=0 ?

a. y = Cx^-4





b. y = Cx^-1/4





c. y = Cx^1/4





d. y = Cx^4





e. sin y + cos x = C





f. cos y + sin x = C





g. cos y - sin x = C





h. sin y - cos x = C





i. none of these

C is an arbitrary constant, from the choices which would be a solution to the differential equation xy'+4y=0 ?
The answer is a because:





xy'+4y=0





y'=-4y/x





since y'=(dy/dx)





we separate variables to solve





(1/y)dy=(-4/x)dx





integrate and get





ln(y)=-4ln(x)+C





y=e^(-4ln(x)+C)





y=Ce^(ln(x^-4)) - Property of logs





e^(ln(x))= x so:





y=Cx^(-4) (A)
Reply:x(dy/dx) +4y =0





xdy/dx = -4y





dy /-4y = dx /x





-4logy = logx+ logc


y^-4 = xc


y = cx^-1/4





potion b





log c
Reply:y´/y= =-4/x


lnI yI =-4lnIxI+K


so y = Cx^-4 (a)