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Second differential equations

  1. Jan 1, 2004 #1
    will you please help
    v" + ë^2v = [-WL/(2EI)]x + [W/(2EI)]x^2
    where ë^2= PE/I
    P, E, I, W are constants;
    how will the general solution of this equation be
    v(x)=Acosëx + Bsinëx + (W/2P)*((x^2)-Lx-2/ë^2)
  2. jcsd
  3. Jan 2, 2004 #2


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    This is a linear, non-homogeneous, differential equation with constant coefficients. A standard way of solving these is to first look at the corresponding homogeneous equation. In this case that is
    v" + ë2v = 0. Such equations typically have exponential (or related)solutions. If you try v= ekx here, v'= k ekx and v"= k2ekx so the equation becomes (k2e+ ë2)ekx= 0. Since an exponential is never 0, we must have k2+ ë2= 0. The solutions to that equation are k= ëi and -ëi. Two independent solutions to the differential equation are eëix and e-ëix but it is simpler to use eëix= cos(ëx)+ i sin(ëx).
    The general solution to the homogeneous equation is
    v(x)= C1cos(ëx)+ C2 sin(ëx).

    Now "look for" a solution to the entire equation- Since the "right hand side", [-WL/(2EI)]x + [W/(2EI)]x2, is a quadratic, try a solution of the form v(x)= Ax2+ Bx+ C. v'= 2Ax+ B and v"= 2A. The equation becomes 2A+ ë2(Ax2+ Bx+ C)= ë2Ax2+ ë2Bx+ 2A+ ë2C= [-WL/(2EI)]x + [W/(2EI)]x2. Equating coefficients of like powers, ë2A= W/(2EI), ë2B= -WL/(2EI), and 2A+ ë2C= 0.

    After solving for A, B, C, add that to the solution to the homogeneous equation: v(x)= C1cos(ëx)+ C2 sin(ëx)+ Ax2+ Bx+ C, the solution you give.

    Solving linear differential equations with constant coefficients is covered in any basic text on differential equations.
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