Answer to Question #289722 in Differential Equations for Loki

Question #289722

(D ^ 2 + 3D + 2) * y = sin 2x + 5 degrees + log 3




1
Expert's answer
2022-01-28T14:42:18-0500

The given question is;

"\\displaystyle\n(D^2+3D+2)y=\\sin(2x+5)+\\log3"

The Auxiliary equation of the given DE in terms of variable m is;

"\\displaystyle\nm^2+3m+2=0\\\\"

"\\displaystyle\n\\Rightarrow(m+1)(m+2)=0\\\\\n\\Rightarrow m=-1,-2\\\\"

Hence, the complementary function is;

"\\displaystyle\ny_c=c_1e^{-x}+c_2e^{-2x}", where "\\displaystyle\nc_1" and "\\displaystyle\nc_2" are arbitrary constants.


Next, to calculate the particular integral "\\displaystyle\n(y_p)" we proceed as follows;

"\\displaystyle\ny_p=\\frac{1}{D^2+3D+2}\\sin2\\left(x+\\frac{5}{2}\\right)+\\frac{1}{D^2+3D+2}(\\log3)\\\\\n\\quad=\\frac{1}{-4+3D+2}\\sin2\\left(x+\\frac{5}{2}\\right)+\\log3\\frac{1}{D^2+3D+2}e^{0x}\\\\\n\\quad=\\frac{1}{3D-2}\\sin2\\left(x+\\frac{5}{2}\\right)+\\frac{\\log3}{2}\\\\\n\\quad=\\frac{3D+2}{9D^2-4}\\sin2\\left(x+\\frac{5}{2}\\right)+\\frac{\\log3}{2}\\\\\n\\quad=\\frac{3D+2}{-40}\\sin2\\left(x+\\frac{5}{2}\\right)+\\frac{\\log3}{2}\\\\\n\\quad=\\frac{-1}{40}\\left[3D\\left(\\sin(2x+5)\\right)+2\\sin(2x+5)\\right]+\\frac{\\log3}{2}\\\\\n\\quad=\\frac{-1}{40}[6\\cos(2x+5)+2\\sin(2x+5)]+\\frac{\\log3}{2}\\\\\n\\quad=-\\frac{3\\cos(2x+5)}{20}-\\frac{\\sin(2x+5)}{20}+\\frac{\\log3}{2}"


Hence, the general solution is;

"\\displaystyle\ny=y_c+y_p\\\\\n\\quad=c_1e^{-x}+c_2e^{-2x}-\\frac{3\\cos(2x+5)}{20}-\\frac{\\sin(2x+5)}{20}+\\frac{\\log3}{2}"


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