Answer to Question #225222 in Differential Equations for Asif

Question

Determine the general solution of 8d²y/dx² - 2dy/dx + 8y = cos(8x) + 8x³ - 3x.

Answer 1

$\textsf{First, we have to find the complementary function.}\\ \textsf{To do this, we sole the equation:}\\ 8\frac{d^2y}{dx^2}-2\frac{dy}{dx}+8y=0\\ \textsf{The auxiliary equation is:}\\ 8m^2-2m+8=0\\ m=\frac{1}{8}\pm\frac{3\sqrt7}{8}i\\ \textsf{Hence, the complementary function, } \\ y_c=e^{\frac x8}(c_1cos(\frac{3\sqrt7x}{8})+c_2sin(\frac{3\sqrt7x}{8}))\\ \textsf{To find the particular integral}, y_p,\\ \textsf{ we assume the general form of the RHS}\\ y_p=Ccos(8x) + Dsin(8x)+Ex^3+Fx^2+Gx+H\\ \frac{dy}{dx}=-8Csin(8x)+8Dcos(8x)+3Ex^2+2Fx+G\\ \frac{d^2y}{dx^2}=-64Ccos(8x)-64Dsin(8x)+6Ex+2F\\ \textsf{Substituting these into the given equation, we have,}\\ 8(-64Ccos(8x)-64Dsin(8x)+6Ex+2F)\\ -2(-8Csin(8x)+8Dcos(8x)+3Ex^2+2Fx+G)\\ +8(Ccos(8x) + Dsin(8x)+Ex^3+Fx^2+Gx+H)\\ =cos(8x)+8x^3-3x\\ (-504C-16D)cos(8x)+(-504D+16C)sin(8x)\\ +8Ex^3+(8F-6E)x^2+(8G-4F+48E)x\\+8H-2G+16F\\ =cos(8x)+8x^3-3x\\ \textsf{By comparing coefficients on both sides,} -504C-16D =1\\ 16C-504D =0\\ 8E=8\\ 8F-6E=0\\ 8G-4F+48E=-3\\ 8H-2G+16F=0\\ \textsf{By solving the equations simultaneously, we have,}\\ C=\frac{-63}{31784},D=\frac{-1}{15892},E=1, F=\frac34,G=-6,H=-3\\ \textsf{Hence, } y_p=-\frac{63}{31784}cos(8x)-\frac{1}{15892}sin(8x)+x^3+\frac{3x^2}{4}-6x-3\\ \textsf{The general solution is given by,}\\ y=y_c+y_p\\ y=e^{\frac x8}(c_1cos(\frac{3\sqrt7x}{8})+c_2sin(\frac{3\sqrt7x}{8}))-\frac{63}{31784}cos(8x)-\frac{1}{15892}sin(8x)+x^3+\frac{3x^2}{4}-6x-3\\$