Posted by Kenneth on Monday, December 1, 2008 at 9:10pm.
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, Monday, December 1, 2008 at 9:18pm
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, Monday, December 1, 2008 at 9:25pm
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.
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