y′′−2y′+y = xe^x sin x


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Question Number 105257 by bemath last updated on 27/Jul/20

$y^{″} - 2 y^{'} + y = {x e}^{x} \sin x$
	Answered by john santu last updated on 27/Jul/20
	$homogenous solution λ^2 −2λ+1 = 0 ⇒λ = 1;1 y_h = C_1 e^x +C_2 xe^x let y_1 = e^x →y′_1 =e^x y_2 = xe^x →y′_2 = e^x +xe^x W(y_1 ,y_2 ) = determinant (((e^x xe^x )),((e^x e^x +xe^x )))= e^(2x) u_1 = −∫((xe^x .(xe^x sin x))/e^(2x) ) dx u_1 =−∫ x^2 sin x dx u_1 =−{−x^2 cos x+2x sin x +2 cos x} u_1 =x^2 cos x −2x sin x −2cos x u_2 = ∫((e^x (xe^x sin x))/e^(2x) ) dx = ∫x sin x dx u_2 = −x cos x + sin x particular solution y_p = y_1 u_1 +y_2 u_2 y_p = e^x {x^2 cos x −2x sin x −2cos x} + xe^x {−xcos x + sin x } Y_p = x^2 e^x cos x−2xe^x sin x−2e^x cos x −x^2 e^x cos x+xe^x sin x Y_p = −xe^x sin x−2e^x cos x General solution Y_g = C_1 e^x +C_2 xe^x −xe^x sin x−2e^x cos x (JS ♠⧫)$
	$h o m o g e n o u s s o l u t i o n$ $λ^{2} - 2 λ + 1 = 0 \Rightarrow λ = 1; 1$ $y_{h} = C_{1} e^{x} + C_{2} {x e}^{x}$ $l e t y_{1} = e^{x} \to y_{1}^{'} = e^{x}$ $y_{2} = {x e}^{x} \to y_{2}^{'} = e^{x} + {x e}^{x}$ $W (y_{1}, y_{2}) = \| \begin{matrix} e^{x} {x e}^{x} \\ e^{x} e^{x} + {x e}^{x} \end{matrix} \| = e^{2 x}$ $u_{1} = - \int \frac{{x e}^{x} . ({x e}^{x} \sin x)}{e^{2 x}} d x$ $u_{1} = - \int x^{2} \sin x d x$ $u_{1} = - {- x^{2} \cos x + 2 x \sin x + 2 \cos x}$ $u_{1} = x^{2} \cos x - 2 x \sin x - 2 \cos x$ $u_{2} = \int \frac{e^{x} ({x e}^{x} \sin x)}{e^{2 x}} d x = \int x \sin x d x$ $u_{2} = - x \cos x + \sin x$ $p a r t i c u l a r s o l u t i o n$ $y_{p} = y_{1} u_{1} + y_{2} u_{2}$ $y_{p} = e^{x} {x^{2} \cos x - 2 x \sin x - 2 \cos x}$ $+ {x e}^{x} {- x \cos x + \sin x}$ $Y_{p} = x^{2} e^{x} \cos x - 2 {x e}^{x} \sin x - 2 e^{x} \cos x$ $- x^{2} e^{x} \cos x + {x e}^{x} \sin x$ $Y_{p} = - {x e}^{x} \sin x - 2 e^{x} \cos x$ $G e n e r a l s o l u t i o n$ $Y_{g} = C_{1} e^{x} + C_{2} {x e}^{x} - {x e}^{x} \sin x - 2 e^{x} \cos x$ $(J S ♠ ⧫)$
	Commented by bemath last updated on 27/Jul/20

	$c o o l l$
	Answered by abdomathmax last updated on 27/Jul/20
	$h→r^2 −2r +1 =0 ⇒(r−1)^2 =0 ⇒r=1 ⇒ y =(ax +b)e^x =axe^(x ) +be^x =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((xe^x e^x )),(((x+1)e^x e^x )))=xe^(2x) −(x+1)e^(2x) =−e^(2x) ≠0 W_1 = determinant (((o e^x )),((xe^x sinx e^x )))=−xe^(2x) sinx W_2 = determinant (((xe^x 0)),(((x+1)e^x xe^x sinx)))=x^2 e^(2x) sinx v_1 =∫ (w_1 /w)dx =∫ ((−xe^(2x) sinx)/(−e^(2x) ))dx =∫ xsinx dx =−xcosx +∫ cosx dx =−xcosx +sinx v_2 =∫ (w_2 /w)dx =∫ ((x^2 e^(2x) sinx)/(−e^(2x) ))dx =−∫ x^2 sinx dx =−{ −x^2 cosx +∫ 2x cosx dx} =x^2 cosx−2 ∫ xcosx dx =x^2 cosx −2{ xsinx −∫ sinx dx} =x^2 cosx −2{xsinx +cosx} =(x^2 −2)cosx−2x sinx ⇒ y_p =u_1 v_(1 ) +u_2 v_2 =xe^x (−x cosx+sinx) +e^x { (x^2 −2)cosx −2xsinx} =−x^2 e^x cosx +xe^x sinx+(x^2 −2)e^(x ) cosx−2xe^x sinx =−2 e^x cosx −x e^x sinx the veneral solution is y =y_(h ) +y_p =axe^x +be^(x ) −2e^x cosx−xe^x sinx$
	$h \to r^{2} - 2 r + 1 = 0 \Rightarrow {(r - 1)}^{2} = 0 \Rightarrow r = 1 \Rightarrow$ $y = (ax + b) e^{x} = {axe}^{x} + {be}^{x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} {xe}^{x} e^{x} \\ (x + 1) e^{x} e^{x} \end{matrix} \| = {xe}^{2 x} - (x + 1) e^{2 x} = - e^{2 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{x} \\ {xe}^{x} sinx e^{x} \end{matrix} \| = - {xe}^{2 x} sinx$ $W_{2} = \| \begin{matrix} {xe}^{x} 0 \\ (x + 1) e^{x} {xe}^{x} sinx \end{matrix} \| = x^{2} e^{2 x} sinx$ $v_{1} = \int \frac{w_{1}}{w} dx = \int \frac{- {xe}^{2 x} sinx}{- e^{2 x}} dx = \int xsinx dx$ $= - xcosx + \int cosx dx = - xcosx + sinx$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{x^{2} e^{2 x} sinx}{- e^{2 x}} dx$ $= - \int x^{2} sinx dx = - {- x^{2} cosx + \int 2 x cosx dx}$ $= x^{2} cosx - 2 \int xcosx dx$ $= x^{2} cosx - 2 {xsinx - \int sinx dx}$ $= x^{2} cosx - 2 {xsinx + cosx}$ $= (x^{2} - 2) cosx - 2 x sinx \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = {xe}^{x} (- x cosx + sinx)$ $+ e^{x} {(x^{2} - 2) cosx - 2 xsinx}$ $= - x^{2} e^{x} cosx + {xe}^{x} sinx + (x^{2} - 2) e^{x} cosx - {2 xe}^{x} sinx$ $= - 2 e^{x} cosx - x e^{x} sinx$ $the veneral solution is$ $y = y_{h} + y_{p} = {axe}^{x} + {be}^{x} - {2 e}^{x} cosx - {xe}^{x} sinx$
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