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I am trying to solve the equation of the following type: $$u''(x) = au(x)-b \operatorname{sech}^2(cx),$$ where $a,b,c$ are positive real constants.

It is my first time when I met the equation which includes $\operatorname{sech}^2(x)$ function.

Any hint will be good for me! Thank you in advance!

  • 1
    Apparently, http://m.wolframalpha.com/input/?i=u%27%27%28x%29+%3D+au%28x%29-bsech%5E2%28cx%29&x=0&y=0 – 2017-02-08
  • 0
    do you know method of variation of parameters?? – 2017-02-08

1 Answers 1

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As usual first you need the homogeneous part and then you need a particular solution to the nonhomogeneous part.

Homogeneous, $u'' - au = 0$

Let u = $e^{kx}$ so $u'' = k^2 e^{kx}$

$k^2 - a = $ and $k = \pm \sqrt{a}$

Homogeneous part of the solution is $u_h = C_1e^{\sqrt{a} x} + C_2e^{-\sqrt{a} x}$

Nonhomogeneous part:

You need the hyperbolic trig functions and their derivatives.

Here is a nice neat reference I found online:

http://tutorial.math.lamar.edu/classes/calci/diffhypertrigfcns.aspx

You will see basic definitions at the top and derivatives about halfway down.

We see that forms including $sech(x)$ are:

$d/dx(tanh(x)) = sech^2(x)$

$d/dx(sech(x))= - sec(x)tanh(x)$

Let $u_p(x) = v_1tanh(cx) + v_2sech(cx)$ where u and v are unknown functions.

Then take the first and second derivatives of these and substitute into your problem

$u''x - au(x) = -b(sech^2(cx))$

There will be a LOT of algebra to sort out but you just work through it.