Question

Solve the ODE

y00 + 4y0 + 4y =

e􀀀2x

x3

using the method of variation of parameters

Answer 1

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