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    Hi,

    So the question asked find general solutions of the following differential equation.

    1/x dy/dx - y/x^2= 1/x^3

    Multiplied both sides by x to Get :
    dy/dx - y/x =1/x^2
    I found out that the general solution for an equation (Attached)

    I applied the formula and got y = Ln x / x + c.

    The book got something completely different.

    Thanks
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    (Original post by ChemBoy1)
    Hi,

    So the question asked find general solutions of the following differential equation.

    1/x dy/dx - y/x^2= 1/x^3

    Multiplied both sides by x to Get :
    dy/dx - y/x =1/x^2
    I found out that the general solution for an equation (Attached)

    I applied the formula and got y = Ln x / x + c.

    The book got something completely different.

    Thanks
    1/x dy/dx - y/x2 is the exact derivative of y*{1/x} using the product rule.

    so when you integrate the LHS you get y*{1/x}
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    Thanks for this but even still, wont the above equation still work?
    (Original post by the bear)
    1/x dy/dx - y/x2 is the exact derivative of y*{1/x} using the product rule.

    so when you integrate the LHS you get y*{1/x}
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    (Original post by ChemBoy1)
    Thanks for this but even still, wont the above equation still work?
    it looks like you used an Integrating Factor.... what was it ?

    :holmes:
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    ~I've given up on differntial equations and c4 for that matter
 
 
 
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