y′′−y′−2y=e^(2x) .cos^2 x


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Question Number 115771 by bemath last updated on 28/Sep/20

$y^{″} - y^{'} - 2 y = e^{2 x} . \cos^{2} x$
	Answered by TANMAY PANACEA last updated on 28/Sep/20

	Answered by mathmax by abdo last updated on 28/Sep/20
	$h→r^2 −r−2=0 →Δ =1−4(−2) =9 ⇒r_1 =((1+3)/2)=2 and r_2 =((1−3)/2)=−1 ⇒y_h =ae^(2x) +be^(−x) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((e^(2x) e^(−x) )),((2e^(2x) −e^(−x) )))=−e^x −2e^x =−3e^x ≠0 W_1 = determinant (((o e^(−x) )),((e^(2x) cos^2 x −e^(−x) )))=−e^x cos^2 x W_2 = determinant (((e^(2x) 0)),((2e^(2x) e^(2x) cos^2 x)))=e^(4x) cos^2 x v_1 =∫ (w_1 /w)dx =−∫ ((e^x cos^2 x)/(−3e^x )) =(1/3)∫ ((1+cos(2x))/2)dx =(1/6)x +(1/(12))sin(2x) v_2 =∫ (w_2 /w)dx =∫ ((e^(4x) cos^2 x)/(−3e^x )) dx =−(1/3)∫ e^(3x) (((1+cos(2x))/2))dx =−(1/6) ∫e^(3x) dx −(1/6)∫ e^(3x ) cos(2x)dx but =−(1/(18))e^(3x) −(1/6)Re(∫ e^((3+2i)x) dx) ∫ e^((3+2i)x) dx =(1/(3+2i)) e^((3+2i)x) =((3−2i)/(13)) e^(3x) (cos(2x)+isin(2x)) =(e^(3x) /(13)){ 3cos(2x)+3isin(2x)−2i cos(2x) +2sin(2x)} ⇒ Re(....) =(e^(3x) /(13))( 3cos(2x)+2sin(2x)) ⇒ y_p =u_1 v_1 +u_2 v_2 =e^(2x) ((x/6)+(1/(12))sin(2x))+(e^(2x) /(13))(3cos(2x)+2sin(2x)) the general solution is y =y_h +y_p$
	$h \to r^{2} - r - 2 = 0 \to Δ = 1 - 4 (- 2) = 9 \Rightarrow r_{1} = \frac{1 + 3}{2} = 2 and r_{2} = \frac{1 - 3}{2} = - 1$ $\Rightarrow y_{h} = {ae}^{2 x} + {be}^{- x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{2 x} e^{- x} \\ {2 e}^{2 x} - e^{- x} \end{matrix} \| = - e^{x} - {2 e}^{x} = - {3 e}^{x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{- x} \\ e^{2 x} \cos^{2} x - e^{- x} \end{matrix} \| = - e^{x} \cos^{2} x$ $W_{2} = \| \begin{matrix} e^{2 x} 0 \\ {2 e}^{2 x} e^{2 x} \cos^{2} x \end{matrix} \| = e^{4 x} \cos^{2} x$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{e^{x} \cos^{2} x}{- {3 e}^{x}} = \frac{1}{3} \int \frac{1 + \cos (2 x)}{2} dx = \frac{1}{6} x + \frac{1}{12} \sin (2 x)$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{e^{4 x} \cos^{2} x}{- {3 e}^{x}} dx = - \frac{1}{3} \int e^{3 x} (\frac{1 + \cos (2 x)}{2}) dx$ $= - \frac{1}{6} \int e^{3 x} dx - \frac{1}{6} \int e^{3 x} \cos (2 x) dx but = - \frac{1}{18} e^{3 x} - \frac{1}{6} Re (\int e^{(3 + 2 i) x} dx)$ $\int e^{(3 + 2 i) x} dx = \frac{1}{3 + 2 i} e^{(3 + 2 i) x} = \frac{3 - 2 i}{13} e^{3 x} (\cos (2 x) + isin (2 x))$ $= \frac{e^{3 x}}{13} {3 \cos (2 x) + 3 isin (2 x) - 2 i \cos (2 x) + 2 \sin (2 x)} \Rightarrow$ $Re (. . . .) = \frac{e^{3 x}}{13} (3 \cos (2 x) + 2 \sin (2 x)) \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{2 x} (\frac{x}{6} + \frac{1}{12} \sin (2 x)) + \frac{e^{2 x}}{13} (3 \cos (2 x) + 2 \sin (2 x))$ $the general solution is y = y_{h} + y_{p}$
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