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Is there a better way of solving this first order linear differential equation?

  1. Calculus III

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  2. Differential Equations

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  1. Feb 3, 2012 #1
    3y'+2y-2sin(3x)+2e(-3x)+x3+4=0

    Variables
    x - independent
    y - dependent

    Attempt at a solution
    I rewrote the equation in form dy/dx+P(x)y=Q(x) and used an integrating factor of [itex]\mu(x)[/itex]=ke(2/3)x
    with P(x) = 2/3 and Q(x) = 2sin(3x)-2e(-3x)-x3-4
    Since y(x) = 1/ke(2/3)x∫e(2x/3)(2sin(3x)-2e(-3x)-x3-4)dx

    This integral lead to two integrals that required integration by parts to be solved which was a tedious process for me. Is there a better way solve this differential equation?
     
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2

    LCKurtz

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    An alternative approach would be to solve the constant coefficient homogeneous equation and use undetermined coefficients to find a particular solution of the non-homogeneous equation. I'm guessing it might be a bit less work.
     
  4. Feb 4, 2012 #3
    3y'+2y-2sin(3x)+2exp(-3x)+x^3+4=0
    Solutions of 3Y'+2Y=0 are Y(x)=C*exp(-2x/3)
    A particular solution of 3(y1)'+2(y1)+4=0 obviously is (y1)=-2
    A particular solution of 3(y2)'+2(y2)+x^3=0 is (y2)=a(x^3)+b(x^2)+cx+d ; compute a,b, c, d.
    A particular solution of 3(y3)'+2(y3)+2exp(-3x)=0 is (y3)=a*exp(-3x) ; compute a.
    A particular solution of 3(y4)'+2(y4)-2sin(3x)=0 is (y4)=a*cos(3x)+b*sin(3x) : compute a, b;
    The solutions of the complete EDO are : y(x)=Y(x)+(y1)+(y2)+(y3)+(y4)
     
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