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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