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[Solved]Another second order DE

  1. Oct 10, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I must solve [itex]x^2y''-2y=x[/itex].


    2. Relevant equations
    Not sure. Reduction of order maybe.


    3. The attempt at a solution
    I notice that [itex]y_1=x^2[/itex] is solution to the homogeneous DE [itex]x^2y''-2y=0[/itex].
    I propose another solution of the form [itex]y_2(x)=v(x)y_1(x)[/itex].
    Plugging this y_2 and its second derivative into the original DE, I reach that [itex]v''x^4+4x^3v'=x[/itex].
    I called a new variable [itex]u=v'[/itex] to reduce the order of the DE. So the DE reduces to [itex]u'x^4+4x^3u=x[/itex].
    I then used the integrating factor method to solve the DE. As integrating factor, I reached [itex]e^{x^4}[/itex]. So that [itex]u(x)=\frac{1}{e^{x^4}} \left [ \int{xe^{x^4}dx+C_1} \right ][/itex]. I tried to evaluate this integral via Wolfram Alpha and it returned the imaginary erf function so I prefer to let it under this form.
    So now my [itex]v(x)=\int u(x)dx+C_2[/itex] which looks extremely "ugly".

    I just tried to solve the original DE into Wolfram and it gives the solution [itex]y(x)=c_1x^2+\frac{c_2}{x}-\frac{x}{2}[/itex]. I really don't know what I did wrong. (In http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx it seems that the original DE must be homogeneous in order to apply the reduction of order method; however it seems I could do the same even though the DE isn't homogeneous. Could you clarify if I'm right on this?)
    Did I do something wrong? How could I reach a "beautiful" result such the one given by Wolfram Alpha?


    EDIT: Nevermind, I found out my error. I reach exactly the same answer as Wolfram Alpha. It was when I applied the integrating factor method, I forgot to divide the DE by x⁴ before doing the math. Nice!
     
    Last edited: Oct 10, 2011
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