Answer to Question #140347 in Differential Equations for Stefanus Amen

Question

Solve the ODE

y00 + 4y0 + 4y =

e􀀀2x

x3

using the method of variation of parameters

Answer 1

"\\displaystyle \n\ny" + 4y' + 4y = x^3 e^{2x}\\\\\n\n\\textsf{The solution to the above equation is}\\\\\ny = y_c + y_p \\\\\n\n\\textsf{where}\\, y_c \\, \\textsf{is the complementary factor and}\\\\\n y_p\\, \\textsf{ is the particular integral}\\\\\n\n\\textsf{The auxiliary equation is} \\,m^2 + 4m + 4 = 0 \\\\\n\n\\begin{aligned}\n(m + 2)(m + 2) &= 0 \\\\\n\\therefore m &= -2 \\, \\textsf{twice}\n\\end{aligned}\\\\\n\\therefore y_c = (C_1x + C_2)e^{-2x}\\\\\n\n\n\\textsf{The Wronskian of the two solution is}\\\\\n\nW(x) = e^{-\\int 4\\, \\mathrm{d}x} = e^{-4x}\\\\\n\ny_p = V_1(x)e^{-2x} + xV_2(x)e^{-2x} \\\\\n\n\n\\begin{aligned}\nV_2(x) &= \\int e^{-2x}\\cdot\\frac{x^3 e^{2x}}{e^{-4x}} = \\int x^3 e^{4x}\\, \\mathrm{d}x\\\\\n&= \\int x^3\\mathrm{d}\\left(\\frac{e^{4x}}{4}\\right)\\\\ &= \\frac{-x^3e^{4x}}{4} + 3\\int\\frac{x^2e^{4x}}{4}\\, \\mathrm{d}x \\\\\n&=\\frac{x^3e^{4x}}{4} - \\frac{3}{4}\\int x^2e^{4x} \\mathrm{d}x\\\\\n&=\\frac{x^3e^{4x}}{4} - \\frac{3}{4}\\int x^2 \\mathrm{d}\\left(\\frac{e^{4x}}{4}\\right)\\\\&=\\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6}{16}\\int xe^{4x}\\, \\mathrm{d}x\\\\&=\\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6}{16}\\int x\\mathrm{d}\\left(\\frac{e^{4x}}{4}\\right)\\\\&=\\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6xe^{4x}}{64} - \\frac{6}{64}\\int e^{4x}\\, \\mathrm{d}x\n\\\\&=\\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6xe^{4x}}{64} - \\frac{6e^{4x}}{256}\n\\end{aligned}\\\\\n\n\n\\begin{aligned}\nV_1(x) &= -\\int xe^{-2x}\\cdot\\frac{x^3 e^{2x}}{e^{-4x}} = -\\int x^4 e^{4x}\\, \\mathrm{d}x\\\\\n&= -\\int x^4\\mathrm{d}\\left(\\frac{e^{4x}}{4}\\right)\\\\ &= -\\frac{x^4e^{4x}}{4} + \\int x^3 e^{4x}\\, \\mathrm{d}x\\\\\n\\\\ &= -\\frac{x^4e^{4x}}{4} + V_2(x)\\\\\n&= -\\frac{x^4e^{4x}}{4} + \\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6xe^{4x}}{64} - \\frac{6e^{4x}}{256}\n\\end{aligned}\\\\\n\n\n\n\\begin{aligned}\ny_p &= xe^{-2x}\\left(\\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6xe^{4x}}{64} - \\frac{6e^{4x}}{256}\\right) \\\\&+ e^{-2x}\\left(\\frac{x^4e^{4x}}{4} + \\frac{x^3e^{4x}}{4} - \\frac{3x^2e^{4x}}{16} + \\frac{6xe^{4x}}{64} - \\frac{6e^{4x}}{256}\\right)\\\\ &=\\frac{x^4e^{2x}}{4} - \\frac{3x^3e^{2x}}{16} + \\frac{6x^2e^{2x}}{64} - \\frac{6xe^{2x}}{256} \\\\&- \\frac{x^4e^{2x}}{4} + \\frac{x^3e^{2x}}{4} - \\frac{3x^2e^{2x}}{16} + \\frac{6xe^{2x}}{64} - \\frac{6e^{2x}}{256}\n\\\\&=\\frac{x^3e^{2x}}{16} - \\frac{24x^2e^{2x}}{256} + \\frac{18xe^{2x}}{256} - \\frac{6e^{2x}}{256}\n\\\\&=\\frac{x^3e^{2x}}{16} - \\frac{3x^2e^{2x}}{32} + \\frac{9xe^{2x}}{128} - \\frac{3e^{2x}}{128}\n\\\\&= \\frac{e^{2x}}{128}\\left(8x^3 - 4x^2 + 9xe^{2x} - 3e^{2x}\\right)\n\\end{aligned}\\\\\n\n\n\\therefore y = (C_1x + C_2)e^{-2x} + \\frac{e^{2x}}{128}\\left(8x^3 - 4x^2 + 9xe^{2x} - 3e^{2x}\\right)"