Question:

Diferencial equation?

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i need help with this lineal diferencial equation. i dont know how to use the integrating factor.

the equation is: x^2 *y' +xy = x

the answer is y = x^-1 lnx +cx^-1

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  1. Rearrange this in the form y'+yp(x)=q(x)

    divide by x^2

    y'+y/x = 1/x

    p(x)=1/x

    q(x)=1/x

    u(x)= e^ ∫ p(x) dx

    the solution is y= ∫ u(x) q(x) dx / u(x) +c




  2. First, divide through by the coefficient of y' to get

    y'+(1/x)y=1/x.

    The integrating factor is

    exp(int 1/x dx)=exp(ln x)=x, so we have

    xy'+y=1.

    Rewrite the left hand side

    (xy)'=1

    And integrate:

    xy=x+C, so

    y=1+c/x.

    This differs from your answer. Are you sure the original problem wasn't

    x^2 y'+xy=1? This would have given the stated answer.

  3. I think here, that first it's a good idea to divide everything by x, so that you get xy' + y = 1

    Then, note that you don't actually need an integrating factor - the x itself acts as one, so that you can rewrite this as

    d/dx(xy) = 1.  

    Then,

    xy = x + c

    y = 1 + c/x

    Note that this is not the same solution as you had, but it does work.
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