Answer to Question #145391 in Differential Equations for Jisha

Question #145391
Solve : (D^2-2D+4) y=8x^2e^2x sin2x
1
Expert's answer
2020-11-19T15:56:00-0500

"(D^2 - 2D + 4)y = 8x^2e^{2x}\\sin{2x}\n\n\n\\\\ \\textsf{The solution to the above equation is}\n\\\\ y = y_c + y_p \\\\\n\\textsf{where}\\, y_c \\, \\textsf{is the complementary factor and}\n\\\\ y_p\\, \\textsf{ is the particular integral}\n\n\n\\textsf{The auxiliary equation to this ODE is}\\\\\n\\begin{aligned}\nm^2 - 2m + 4 &= 0\\\\\nm &= \\frac{2 \\pm \\sqrt{4 - 16}}{2} = \\frac{2 \\pm j\\sqrt{12}}{2}\n\\\\&=\\frac{2 \\pm j2\\sqrt{3}}{2}= 1 \\pm j\\sqrt{3}\n\\end{aligned}\\\\\n\n\n\\textsf{The complementary solution to this ODE is}\\\\\n\ny = e^{x}(A\\cos(\\sqrt{3}x) + B\\sin(\\sqrt{3}))\\\\\n\n\\textsf{The Wronskian of the two solutions is} \\\\\n\nW(x) = e^{-\\int-2\\mathrm{d}x} = e^{2x}\\\\\n\n\\therefore\\textsf{ Our particular solution will be given by}\\\\ \ny_p = V_1(x)e^{x}\\sin(\\sqrt{3}x) + V_2(x)e^{x}\\cos(\\sqrt{3}x)\\\\\n\n\\textsf{Where}\\, V_1(x) = -\\int \\frac{r(x)\\sin(\\sqrt{3}x)}{W(x)} \\, \\mathrm{d}x,\\, V_2(x) = \\int \\frac{r(x)\\cos(\\sqrt{3}x)}{W(x)}\\, \\mathrm{d}x \\, \\textsf{and} \\, \\\\\n\n\n\n\\begin{aligned}\nV_1(x) &= -\\int \\frac{8x^2e^{2x}\\sin{2x} \\sin(\\sqrt{3}x)}{e^{2x}} \\, \\mathrm{d}x\\\\\n&= -\\int 8x^2\\sin{2x} \\sin(\\sqrt{3}x) \\, \\mathrm{d}x\\\\\n&= - 8\\sqrt{3} (8 x \\sin^2(x) - (x^2 - 30) \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n& + 16 \\cos^2(x) ((x^2 - 26) \\cos(\\sqrt{3} x) + 4 \\sqrt{3} x \\sin(\\sqrt{3} x)) \\\\\n&- 16 ((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\cos(\\sqrt{3} x) + C\n\\end{aligned}\\\\\n\n\\begin{aligned}\nV_2(x) &= \\int \\frac{8x^2e^{2x}\\sin{2x} \\cos(\\sqrt{3}x)}{e^{2x}} \\, \\mathrm{d}x\\\\\n&= \\int 8x^2\\sin{2x} \\cos(\\sqrt{3}x) \\, \\mathrm{d}x\\\\\n&= 16((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n&+ 16\\cos^2(x) (4 \\sqrt{3} x \\cos(\\sqrt{3} x) - (x^2 - 26) \\sin(\\sqrt{3} x)) \\\\\n&+ 16\\sqrt{3} (x^2 - 30) \\sin(x) \\cos(\\sqrt{3} x) \\cos(x) \\\\\n&- 64 \\sqrt{3} x \\sin^2(x) \\cos(\\sqrt{3} x)) + C\n\\end{aligned}\\\\\n \n\n\\begin{aligned}\ny_p &= e^{x}\\sin(\\sqrt{3}x)(- 8\\sqrt{3} (8 x \\sin^2(x) - (x^2 - 30) \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n& + 16 \\cos^2(x) ((x^2 - 26) \\cos(\\sqrt{3} x) + 4 \\sqrt{3} x \\sin(\\sqrt{3} x)) \\\\\n&- 16 ((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\cos(\\sqrt{3} x)) \\\\\n&+ e^{x}\\cos(\\sqrt{3}x)(16((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n&+ 16\\cos^2(x) (4 \\sqrt{3} x \\cos(\\sqrt{3} x) - (x^2 - 26) \\sin(\\sqrt{3} x)) \\\\\n&+ 16\\sqrt{3} (x^2 - 30) \\sin(x) \\cos(\\sqrt{3} x) \\cos(x) \\\\\n&- 64 \\sqrt{3} x \\sin^2(x) \\cos(\\sqrt{3} x))\n\\end{aligned}\\\\\n\n\n\n\n\n\\begin{aligned}\n\\therefore y &= e^{x}(A\\cos(\\sqrt{3}x) + B\\sin(\\sqrt{3})) \n\\\\&+ e^{x}\\sin(\\sqrt{3}x)(- 8\\sqrt{3} (8 x \\sin^2(x) - (x^2 - 30) \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n& + 16 \\cos^2(x) ((x^2 - 26) \\cos(\\sqrt{3} x) + 4 \\sqrt{3} x \\sin(\\sqrt{3} x)) \\\\\n&- 16 ((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\cos(\\sqrt{3} x)) \\\\\n&+ e^{x}\\cos(\\sqrt{3}x)(16((x^2 - 26) \\sin^2(x) + 7 x \\sin(2 x)) \\sin(\\sqrt{3} x) \\\\\n&+ 16\\cos^2(x) (4 \\sqrt{3} x \\cos(\\sqrt{3} x) - (x^2 - 26) \\sin(\\sqrt{3} x)) \\\\\n&+ 16\\sqrt{3} (x^2 - 30) \\sin(x) \\cos(\\sqrt{3} x) \\cos(x) \\\\\n&- 64 \\sqrt{3} x \\sin^2(x) \\cos(\\sqrt{3} x))\n\\end{aligned}\\\\"


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