calculus
posted by Kenneth .
is y = x^3 a solution to the differential equation xy'3y=0??
how do i go about solving this??? also, is there a trick to understanding differential equations? i'm really struggling with this idea, but i'm too embarassed to ask my professor for help.

To find out if your answer is correct, differentiate it and see if you get the original back.
y = x^3
dy/dx = 3 x^2
now put that in d dy/dx  3 y
x(3 x^2) 3 x^3 = ?
3 x^3  3 x^3 = 0
sure enough it works 
Well, here try getting all the y stuff on one side and all the x stuff on the other side (separating variables)
x dy/dx = 3 y
dy/y = 3 dx/x
integrate both sides
ln y = 3 ln x
but 3 ln x = ln x^3
so
ln y = ln x^3
so
y = x^3 will work.