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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 -

    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 -

    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
    ln y = ln x^3
    y = x^3 will work.

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