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Question Number 132639 by Ahmed1hamouda last updated on 15/Feb/21

Commented by Ahmed1hamouda last updated on 15/Feb/21

$solve the differential equation$
	Answered by mathmax by abdo last updated on 15/Feb/21
	$h→y^((2)) +2y^((1)) +5y =0 →r^2 +2r+5=0 Δ^′ =1−5=−4 ⇒r_1 =−1+2i and r_2 =−1−2i ⇒ y_h =ae^((−1+2i)x) +b e^((−1−2i)x) =e^(−x) {αcos(2x)+βsin(2x)} =αu_1 +βu_2 W(u_1 ,u_2 )= determinant (((e^(−x) cos(2x) e^(−x) sin(2x))),(((−cos(2x)−2sin(2x))e^(−x) (−sin(2x)+2cos(2x))e^(−x) ))) =e^(−2x) {−cos(2x)sin(2x)+2cos^2 (2x)}−e^(−2x) {−cos(2x)sin(2x)−2sin^2 (2x)} =2e^(−2x) ≠0 W_1 = determinant (((o e^(−x) sin(2x))),((6e^(2x) +xsin^2 x+e^x cos(2x) (−sin(2x)+2cos(2x))e^(−x) ))) =−e^(−x) sin(2x){6e^(2x) +xsin^2 x +e^x cos(2x)} =6e^x sin(2x)−xe^(−x) sin(2x)sin^2 x−sin(2x).cos(2x) W_2 = determinant (((e^(−x) cos(2x) 0)),(((−cos(2x)−2sin(2x))e^(−x) 6e^(2x) +xsin^2 x +e^x cos(2x)))) =6e^x cos(2x)+xe^(−x) cos(2x)sin^2 x +cos^2 (2x) V_1 =∫ (W_1 /W)dx =(1/2)∫ e^(2x) {6e^x sin(2x)−xe^(−x) sin(2x)sin^2 x−(1/2)sin(4x)}dx =3∫ e^(3x) sin(2x)dx−(1/2)∫xe^x sin(2x)sin^2 x dx−(1/4)∫e^(2x) sin(4x)dx ∫ e^(3x) sin(2x)dx =Im(∫ e^(3x+2ix) dx)=.... ∫ e^(2x) sin(4x)dx =Im(∫ e^(2x+4ix) dx)=... ∫ xe^x sin(2x)sin^2 (x)dx =∫ xe^x sin(2x)((1−cos(2x))/2)dx =(1/2)∫ xe^x sin(2x)dx−(1/4)∫ xe^x sin(4x)dx = =(1/2)Im(∫ x e^(x+2ix) dx)−(1/4)Im(∫ xe^(x+4ix) dx)=... V_2 =∫(W_2 /W)dx =(1/2)∫ e^(2x) {6e^x cos(2x)+xe^(−x) cos(2x)sin^2 x+cos^2 (2x)} =.... ⇒y_p =u_1 v_1 +u_2 v_2 and general solution is y =y_p +y_h$
	$h \to y^{(2)} + {2 y}^{(1)} + 5 y = 0 \to r^{2} + 2 r + 5 = 0$ $Δ^{^{'}} = 1 - 5 = - 4 \Rightarrow r_{1} = - 1 + 2 i and r_{2} = - 1 - 2 i \Rightarrow$ $y_{h} = {ae}^{(- 1 + 2 i) x} + b e^{(- 1 - 2 i) x} = e^{- x} {α \cos (2 x) + β \sin (2 x)}$ $= α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{- x} \cos (2 x) e^{- x} \sin (2 x) \\ (- \cos (2 x) - 2 \sin (2 x)) e^{- x} (- \sin (2 x) + 2 \cos (2 x)) e^{- x} \end{matrix} \|$ $= e^{- 2 x} {- \cos (2 x) \sin (2 x) + {2 \cos}^{2} (2 x)} - e^{- 2 x} {- \cos (2 x) \sin (2 x) - {2 \sin}^{2} (2 x)}$ $= {2 e}^{- 2 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{- x} \sin (2 x) \\ {6 e}^{2 x} + {xsin}^{2} x + e^{x} \cos (2 x) (- \sin (2 x) + 2 \cos (2 x)) e^{- x} \end{matrix} \|$ $= - e^{- x} \sin (2 x) {{6 e}^{2 x} + {xsin}^{2} x + e^{x} \cos (2 x)}$ $= {6 e}^{x} \sin (2 x) - {xe}^{- x} \sin (2 x) \sin^{2} x - \sin (2 x) . \cos (2 x)$ $W_{2} = \| \begin{matrix} e^{- x} \cos (2 x) 0 \\ (- \cos (2 x) - 2 \sin (2 x)) e^{- x} {6 e}^{2 x} + {xsin}^{2} x + e^{x} \cos (2 x) \end{matrix} \|$ $= {6 e}^{x} \cos (2 x) + {xe}^{- x} \cos (2 x) \sin^{2} x + \cos^{2} (2 x)$ $V_{1} = \int \frac{W_{1}}{W} dx = \frac{1}{2} \int e^{2 x} {{6 e}^{x} \sin (2 x) - {xe}^{- x} \sin (2 x) \sin^{2} x - \frac{1}{2} \sin (4 x)} dx$ $= 3 \int e^{3 x} \sin (2 x) dx - \frac{1}{2} \int {xe}^{x} \sin (2 x) \sin^{2} x dx - \frac{1}{4} \int e^{2 x} \sin (4 x) dx$ $\int e^{3 x} \sin (2 x) dx = Im (\int e^{3 x + 2 ix} dx) = . . . .$ $\int e^{2 x} \sin (4 x) dx = Im (\int e^{2 x + 4 ix} dx) = . . .$ $\int {xe}^{x} \sin (2 x) \sin^{2} (x) dx = \int {xe}^{x} \sin (2 x) \frac{1 - \cos (2 x)}{2} dx$ $= \frac{1}{2} \int {xe}^{x} \sin (2 x) dx - \frac{1}{4} \int {xe}^{x} \sin (4 x) dx =$ $= \frac{1}{2} Im (\int x e^{x + 2 ix} dx) - \frac{1}{4} Im (\int {xe}^{x + 4 ix} dx) = . . .$ $V_{2} = \int \frac{W_{2}}{W} dx = \frac{1}{2} \int e^{2 x} {{6 e}^{x} \cos (2 x) + {xe}^{- x} \cos (2 x) \sin^{2} x + \cos^{2} (2 x)}$ $= . . . . \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is$ $y = y_{p} + y_{h}$
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