Help with a linear differential equation

[Sagar_C]

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.

 

[HallsofIvy]

That's a linear equation with constant coefficients. it characteristic equation is D^6+a D^4+b D^2= 0 (we are now thinking of "D" as a number, not a derivative operator). You can obviously factor out D^2 leaving D^4+ aD^2+ b= 0, a quadratic equation in D^2 so, using the quadratic formula, D^2= \frac{-a\pm\sqrt{a^2- 4b}}{2} 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 e^{ax}(A cos(bx)+ B sin(bx)) 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

 

[Sagar_C]

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. :(

 

[HallsofIvy]

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.