Question

Solve : (D^2-2D+4) y=8x^2e^2x sin2x

Answer 1

$(D^2 - 2D + 4)y = 8x^2e^{2x}\sin{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 to this ODE is}\\ \begin{aligned} m^2 - 2m + 4 &= 0\\ m &= \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm j\sqrt{12}}{2} \\&=\frac{2 \pm j2\sqrt{3}}{2}= 1 \pm j\sqrt{3} \end{aligned}\\ \textsf{The complementary solution to this ODE is}\\ y = e^{x}(A\cos(\sqrt{3}x) + B\sin(\sqrt{3}))\\ \textsf{The Wronskian of the two solutions is} \\ W(x) = e^{-\int-2\mathrm{d}x} = e^{2x}\\ \therefore\textsf{ Our particular solution will be given by}\\ y_p = V_1(x)e^{x}\sin(\sqrt{3}x) + V_2(x)e^{x}\cos(\sqrt{3}x)\\ \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} \, \\ \begin{aligned} V_1(x) &= -\int \frac{8x^2e^{2x}\sin{2x} \sin(\sqrt{3}x)}{e^{2x}} \, \mathrm{d}x\\ &= -\int 8x^2\sin{2x} \sin(\sqrt{3}x) \, \mathrm{d}x\\ &= - 8\sqrt{3} (8 x \sin^2(x) - (x^2 - 30) \sin(2 x)) \sin(\sqrt{3} x) \\ & + 16 \cos^2(x) ((x^2 - 26) \cos(\sqrt{3} x) + 4 \sqrt{3} x \sin(\sqrt{3} x)) \\ &- 16 ((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \cos(\sqrt{3} x) + C \end{aligned}\\ \begin{aligned} V_2(x) &= \int \frac{8x^2e^{2x}\sin{2x} \cos(\sqrt{3}x)}{e^{2x}} \, \mathrm{d}x\\ &= \int 8x^2\sin{2x} \cos(\sqrt{3}x) \, \mathrm{d}x\\ &= 16((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \sin(\sqrt{3} x) \\ &+ 16\cos^2(x) (4 \sqrt{3} x \cos(\sqrt{3} x) - (x^2 - 26) \sin(\sqrt{3} x)) \\ &+ 16\sqrt{3} (x^2 - 30) \sin(x) \cos(\sqrt{3} x) \cos(x) \\ &- 64 \sqrt{3} x \sin^2(x) \cos(\sqrt{3} x)) + C \end{aligned}\\ \begin{aligned} y_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) \\ & + 16 \cos^2(x) ((x^2 - 26) \cos(\sqrt{3} x) + 4 \sqrt{3} x \sin(\sqrt{3} x)) \\ &- 16 ((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \cos(\sqrt{3} x)) \\ &+ e^{x}\cos(\sqrt{3}x)(16((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \sin(\sqrt{3} x) \\ &+ 16\cos^2(x) (4 \sqrt{3} x \cos(\sqrt{3} x) - (x^2 - 26) \sin(\sqrt{3} x)) \\ &+ 16\sqrt{3} (x^2 - 30) \sin(x) \cos(\sqrt{3} x) \cos(x) \\ &- 64 \sqrt{3} x \sin^2(x) \cos(\sqrt{3} x)) \end{aligned}\\ \begin{aligned} \therefore y &= e^{x}(A\cos(\sqrt{3}x) + B\sin(\sqrt{3})) \\&+ e^{x}\sin(\sqrt{3}x)(- 8\sqrt{3} (8 x \sin^2(x) - (x^2 - 30) \sin(2 x)) \sin(\sqrt{3} x) \\ & + 16 \cos^2(x) ((x^2 - 26) \cos(\sqrt{3} x) + 4 \sqrt{3} x \sin(\sqrt{3} x)) \\ &- 16 ((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \cos(\sqrt{3} x)) \\ &+ e^{x}\cos(\sqrt{3}x)(16((x^2 - 26) \sin^2(x) + 7 x \sin(2 x)) \sin(\sqrt{3} x) \\ &+ 16\cos^2(x) (4 \sqrt{3} x \cos(\sqrt{3} x) - (x^2 - 26) \sin(\sqrt{3} x)) \\ &+ 16\sqrt{3} (x^2 - 30) \sin(x) \cos(\sqrt{3} x) \cos(x) \\ &- 64 \sqrt{3} x \sin^2(x) \cos(\sqrt{3} x)) \end{aligned}\\$