(D^2 −4D)y = x^2 e^(2x)


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Question Number 103138 by bemath last updated on 13/Jul/20

$(D^{2} - 4 D) y = x^{2} e^{2 x}$
Commented by bemath last updated on 14/Jul/20

Commented by Israrchajk724 last updated on 13/Jul/20

$please any one tell me can we convert math editor to word . . . ?$
	Answered by mathmax by abdo last updated on 14/Jul/20
	$y^(′′) −4y^′ =x^2 e^(2x) (h)→r^2 −4r =0 ⇒r(r−4)=0 ⇒r =o or r=4 ⇒y_h =a +be^(4x) =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((1 e^(4x) )),((0 4e^(4x) )))=4e^(4x) ≠0 W_1 = determinant (((o e^(4x) )),((x^2 e^(2x) 4e^(4x) )))=−x^2 e^(6x) W_2 = determinant (((1 o)),((1 x^2 e^(2x) )))=x^2 e^(2x) v_1 =∫ (w_1 /W)dx =∫ ((−x^2 e^(6x) )/(4e^(4x) ))dx =−(1/4)∫x^2 e^(2x) dx ∫x^2 e^(2x) dx =(x^2 /2)e^(2x) −∫ 2x×(1/2)e^(2x) dx =(x^2 /2)e^(2x) −∫ xe^(2x) dx =(x^2 /2)e^(2x) −{ (x/2) e^(2x) −∫ (1/2)e^(2x) dx} =(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) ⇒ v_1 =−(1/4){(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) } v_2 =∫ (w_2 /w)dx =∫ ((x^2 e^(2x) )/(4e^(4x) ))dx =(1/4) ∫ x^2 e^(−2x) dx 4v_2 =−(x^2 /2)e^(−2x) −∫ 2x×(−(1/2))e^(−2x) dx =−(x^2 /2)e^(−2x) +∫ xe^(−2x) dx =−(x^2 /2)e^(−2x) −(x/2)e^(−2x) +∫(1/2)e^(−2x) dx =−(x^2 /2)e^(−2x) −(x/2)e^(−2x) −(1/4)e^(−2x) ⇒ v_2 =(1/4){ −(x^2 /2)−(x/2)−(1/4)}e^(−2x) y_p =u_1 v_1 +u_2 v_2 =−(1/4){(x^2 /2)−(x/2) +(1/4)}e^(2x) +e^(4x) ×(1/4){−(x^2 /2)−(x/2)−(1/4)}e^(−2x) =(1/4){−(x^2 /2) +(x/2)−(1/4)}e^(2x) +(1/4){−(x^2 /2)−(x/2)−(1/4)}e^(2x) =(1/4){−x^2 −(1/2)}e^(2x) =−(1/4)(x^2 +(1/2))e^(2x) y =y_(h ) +y_p$
	$y^{^{″}} - {4 y}^{^{'}} = x^{2} e^{2 x}$ $(h) \to r^{2} - 4 r = 0 \Rightarrow r (r - 4) = 0 \Rightarrow r = o or r = 4 \Rightarrow y_{h} = a + {be}^{4 x}$ $= {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} 1 e^{4 x} \\ 0 {4 e}^{4 x} \end{matrix} \| = {4 e}^{4 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{4 x} \\ x^{2} e^{2 x} {4 e}^{4 x} \end{matrix} \| = - x^{2} e^{6 x}$ $W_{2} = \| \begin{matrix} 1 o \\ 1 x^{2} e^{2 x} \end{matrix} \| = x^{2} e^{2 x}$ $v_{1} = \int \frac{w_{1}}{W} dx = \int \frac{- x^{2} e^{6 x}}{{4 e}^{4 x}} dx = - \frac{1}{4} \int x^{2} e^{2 x} dx$ $\int x^{2} e^{2 x} dx = \frac{x^{2}}{2} e^{2 x} - \int 2 x \times \frac{1}{2} e^{2 x} dx = \frac{x^{2}}{2} e^{2 x} - \int {xe}^{2 x} dx$ $= \frac{x^{2}}{2} e^{2 x} - {\frac{x}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} dx} = \frac{x^{2}}{2} e^{2 x} - \frac{x}{2} e^{2 x} + \frac{1}{4} e^{2 x} \Rightarrow$ $v_{1} = - \frac{1}{4} {\frac{x^{2}}{2} e^{2 x} - \frac{x}{2} e^{2 x} + \frac{1}{4} e^{2 x}}$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{x^{2} e^{2 x}}{{4 e}^{4 x}} dx = \frac{1}{4} \int x^{2} e^{- 2 x} dx$ ${4 v}_{2} = - \frac{x^{2}}{2} e^{- 2 x} - \int 2 x \times (- \frac{1}{2}) e^{- 2 x} dx = - \frac{x^{2}}{2} e^{- 2 x} + \int {xe}^{- 2 x} dx$ $= - \frac{x^{2}}{2} e^{- 2 x} - \frac{x}{2} e^{- 2 x} + \int \frac{1}{2} e^{- 2 x} dx = - \frac{x^{2}}{2} e^{- 2 x} - \frac{x}{2} e^{- 2 x} - \frac{1}{4} e^{- 2 x} \Rightarrow$ $v_{2} = \frac{1}{4} {- \frac{x^{2}}{2} - \frac{x}{2} - \frac{1}{4}} e^{- 2 x}$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = - \frac{1}{4} {\frac{x^{2}}{2} - \frac{x}{2} + \frac{1}{4}} e^{2 x} + e^{4 x} \times \frac{1}{4} {- \frac{x^{2}}{2} - \frac{x}{2} - \frac{1}{4}} e^{- 2 x}$ $= \frac{1}{4} {- \frac{x^{2}}{2} + \frac{x}{2} - \frac{1}{4}} e^{2 x} + \frac{1}{4} {- \frac{x^{2}}{2} - \frac{x}{2} - \frac{1}{4}} e^{2 x}$ $= \frac{1}{4} {- x^{2} - \frac{1}{2}} e^{2 x} = - \frac{1}{4} (x^{2} + \frac{1}{2}) e^{2 x}$ $y = y_{h} + y_{p}$
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