Posted by Kenneth on .
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.

calculus 
Damon,
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 
calculus 
Damon,
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.