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Solve by using variation of parameters

  1. Nov 19, 2012 #1
    x²y"(x)-3xy'(x)+3y(x)=2(x^4)(e^x)

    =>y"(x)-(3/x)y'(x)+(3/x²)y(x)=2x²e^x

    i dont know how to approach this problem because the coefficients are not constant and i am used to being given y1 and y2

    HELP!!!
     
  2. jcsd
  3. Nov 19, 2012 #2

    HallsofIvy

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    This is an "Euler type" or "equi-potential" equation. The substitution t= ln(x) will change it to a "constant coefficients" problem in the variable t.

    [tex]\frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}[/tex]
    and, differentiating again,
    [tex]\frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)[/tex][tex]= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\right)= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{d^2y}{dt^2}[/tex]
     
  4. Nov 19, 2012 #3
    thank you very much. i appreciate the help
     
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