Help with a linear differential equation

  • Thread starter Sagar_C
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Main Question or Discussion Point

Is there any general method to solve the following linear equation:

[ D^6+a D^4+b D^2+(c-d sech^2(x))] y=0?

Here, D=d/dx and a,b,c,d are constants.
 

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  • #2
HallsofIvy
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That's a linear equation with constant coefficients. it characteristic equation is [itex]D^6+a D^4+b D^2= 0[/itex] (we are now thinking of "D" as a number, not a derivative operator). You can obviously factor out [itex]D^2[/itex] leaving [itex]D^4+ aD^2+ b= 0[/itex], a quadratic equation in [itex]D^2[/itex] so, using the quadratic formula, [itex]D^2= \frac{-a\pm\sqrt{a^2- 4b}}{2}[/itex] and then find D by taking the square root of each of those. The fact that D is a double root of the characteristic equation means that two solutions are a constant and x. For given values of a and b, 0 may b a triple or higher root which would give additional powers of x. For non-zero roots, of the form a+ bi, there will be solutions of the form [itex]e^{ax}(A cos(bx)+ B sin(bx))[/itex] for constants A and B. You will need six independent solutions.

As for the "non-homogeneous" part, try "variation of parameters": http://www.sosmath.com/diffeq/second/variation/variation.html
 
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As for the "non-homogeneous" part, try "variation of parameters": http://www.sosmath.com/diffeq/second/variation/variation.html
Thanks. Just to double-check, so I can treat (c-d sech^2(x))y as the non-homogeneous part even though it contains y? I actually knew how to solve it (as particular solution) when y is absent but not when y is present. :(
 
  • #4
HallsofIvy
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Blast! I didn't see that "y". That is still a linear equation but now with "variable coefficients. Probably the simplest way to handle it is through a power series expansion.
 

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