Determine the general solution of 8d2y/dx2 - 2dy/dx + 8y = cos(8x) + 8x3 - 3x.
"\\textsf{First, we have to find the complementary function.}\\\\\n\\textsf{To do this, we sole the equation:}\\\\\n8\\frac{d^2y}{dx^2}-2\\frac{dy}{dx}+8y=0\\\\\n\\textsf{The auxiliary equation is:}\\\\\n8m^2-2m+8=0\\\\\nm=\\frac{1}{8}\\pm\\frac{3\\sqrt7}{8}i\\\\\n\\textsf{Hence, the complementary function, } \\\\\ny_c=e^{\\frac x8}(c_1cos(\\frac{3\\sqrt7x}{8})+c_2sin(\\frac{3\\sqrt7x}{8}))\\\\\n\\textsf{To find the particular integral}, y_p,\\\\\n\\textsf{ we assume the general form of the RHS}\\\\\ny_p=Ccos(8x) + Dsin(8x)+Ex^3+Fx^2+Gx+H\\\\\n\\frac{dy}{dx}=-8Csin(8x)+8Dcos(8x)+3Ex^2+2Fx+G\\\\\n\\frac{d^2y}{dx^2}=-64Ccos(8x)-64Dsin(8x)+6Ex+2F\\\\\n\\textsf{Substituting these into the given equation, we have,}\\\\\n8(-64Ccos(8x)-64Dsin(8x)+6Ex+2F)\\\\\n-2(-8Csin(8x)+8Dcos(8x)+3Ex^2+2Fx+G)\\\\\n+8(Ccos(8x) + Dsin(8x)+Ex^3+Fx^2+Gx+H)\\\\\n=cos(8x)+8x^3-3x\\\\\n(-504C-16D)cos(8x)+(-504D+16C)sin(8x)\\\\\n+8Ex^3+(8F-6E)x^2+(8G-4F+48E)x\\\\+8H-2G+16F\\\\\n=cos(8x)+8x^3-3x\\\\\n\\textsf{By comparing coefficients on both sides,}\n -504C-16D =1\\\\\n16C-504D =0\\\\\n8E=8\\\\\n8F-6E=0\\\\\n8G-4F+48E=-3\\\\\n8H-2G+16F=0\\\\\n\\textsf{By solving the equations simultaneously, we have,}\\\\\nC=\\frac{-63}{31784},D=\\frac{-1}{15892},E=1, F=\\frac34,G=-6,H=-3\\\\\n\\textsf{Hence, } y_p=-\\frac{63}{31784}cos(8x)-\\frac{1}{15892}sin(8x)+x^3+\\frac{3x^2}{4}-6x-3\\\\\n\\textsf{The general solution is given by,}\\\\\ny=y_c+y_p\\\\\ny=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\\\\"
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