Question

find the form of the general solution of y^(4)(x) - n^2y''(x)=g(x)

Answer 1

The differential equation

$y^{(4)}-n^2y'' = g(x)$

has characteristic equation

r ⁴ - n ² r ² = r ² (r ² - n ²) = r ² (r - n) (r + n) = 0

with roots r = 0 (multiplicity 2), r = -1, and r = 1, so the characteristic solution is

$y_c=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}$

For the non-homogeneous equation, reduce the order by substituting u(x) = y''(x), so that u''(x) is the 4th derivative of y, and

$u''-n^2u = g(x)$

Solve for u by using the method of variation of parameters. Note that the characteristic equation now only admits the two exponential solutions found earlier; I denote them by u₁ and u₂. Now we look for a particular solution of the form

$u_p = u_1z_1 + u_2z_2$

where

$\displaystyle z_1(x) = -\int\frac{u_2(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx$

$\displaystyle z_2(x) = \int\frac{u_1(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx$

where W (u₁, u₂) is the Wronskian of u₁ and u₂. We have

$W(u_1(x),u_2(x)) = \begin{vmatrix}e^{-nx}&e^{nx}\\-ne^{-nx}&ne^{nx}\end{vmatrix} = 2n$

and so

$\displaystyle z_1(x) = -\frac1{2n}\int e^{nx}g(x)\,\mathrm dx$

$\displaystyle z_2(x) = \frac1{2n}\int e^{-nx}g(x)\,\mathrm dx$

So we have

$\displaystyle u_p = -\frac1{2n}e^{-nx}\int_0^x e^{n\xi}g(\xi)\,\mathrm d\xi + \frac1{2n}e^{nx}\int_0^xe^{-n\xi}g(\xi)\,\mathrm d\xi$

and hence

$u(x)=C_1e^{-nx}+C_2e^{nx}+u_p(x)$

Finally, integrate both sides twice to solve for y :

$\displaystyle y(x)=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}+\int_0^x\int_0^\omega u_p(\xi)\,\mathrm d\xi\,\mathrm d\omega$