Solution (d^2 y/dx^2 ) + 3(dy/dx) − 4y = x^2


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Question Number 117728 by syamil last updated on 13/Oct/20

$S o l u t i o n \frac{d^{2} y}{{d x}^{2}} + 3 \frac{d y}{d x} - 4 y = x^{2}$
	Answered by TANMAY PANACEA last updated on 13/Oct/20

	$y = e^{m x}$ $(m^{2} + 3 m - 4) e^{m x} = 0 e^{m x} \neq 0$ $m^{2} + 4 m - m - 4 = 0$ $(m + 4) (m - 1) = 0 \to m = 1, - 4$ $C . F = C_{1} e^{x} + C_{2} e^{- 4 x}$ $P . I = \frac{x^{2}}{D^{2} + 3 D - 4} = \frac{x^{2}}{(D + 4) (D - 1)} = \frac{1}{5} (\frac{1}{D - 1} - \frac{1}{D + 4}) x^{2}$ $y = \frac{1}{5} [\frac{- 1}{1 - D} - \frac{1}{4 (1 + \frac{D}{4})}] x^{2}$ $y = \frac{- 1}{5} [{(1 - D)}^{- 1} + \frac{1}{4} {(1 + \frac{D}{4})}^{- 1}] x^{2}$ $y = \frac{- 1}{5} [(1 + D + D^{2} + D^{3} + . . .) + \frac{1}{4} (1 - \frac{D}{4} + \frac{D^{2}}{16} - \frac{D^{3}}{64})] x^{2}$ $y = \frac{- 1}{5} [(x^{2} + 2 x + 2) + \frac{1}{4} (x^{2} - \frac{2 x}{4} + \frac{2}{16})]$ $= \frac{- 1}{5} [\frac{5 x^{2}}{4} + 2 x (1 - \frac{1}{16}) + 2 + \frac{2}{64}]$ $= \frac{- 1}{5} [\frac{5 x^{2}}{4} + \frac{15 x}{8} + \frac{130}{64}]$ $y = C_{1} e^{x} + C_{2} e^{- 4 x} - \frac{1}{5} (\frac{5 x^{2}}{4} + \frac{15 x}{8} + \frac{65}{32})$
	Answered by john santu last updated on 14/Oct/20
	$particular solution y_p = ax^2 +bx+c → { ((y′=2ax+b)),((y′′=2a)) :} comparing coefficient ⇒2a+3(2ax+b)−4(ax^2 +bx+c)=x^2 −4ax^2 +(6a−4b)x+2a+3b−4c=x^2 { ((−4a=1→a=−(1/4))),((6a−4b=0→b=(3/2)×(−(1/4))=−(3/8))),((2a+3b−4c=0⇒c=(1/4)(−(1/2)−(9/8))=−((13)/(32)))) :} y_p = −(1/4)x^2 −(3/8)x−((13)/(32))$
	$p a r t i c u l a r s o l u t i o n$ $y_{p} = {a x}^{2} + b x + c \to {\begin{cases} y^{'} = 2 a x + b \\ y^{″} = 2 a \end{cases}$ $c o m p a r i n g c o e f f i c i e n t$ $\Rightarrow 2 a + 3 (2 a x + b) - 4 ({a x}^{2} + b x + c) = x^{2}$ $- 4 {a x}^{2} + (6 a - 4 b) x + 2 a + 3 b - 4 c = x^{2}$ ${\begin{cases} - 4 a = 1 \to a = - \frac{1}{4} \\ 6 a - 4 b = 0 \to b = \frac{3}{2} \times (- \frac{1}{4}) = - \frac{3}{8} \\ 2 a + 3 b - 4 c = 0 \Rightarrow c = \frac{1}{4} (- \frac{1}{2} - \frac{9}{8}) = - \frac{13}{32} \end{cases}$ $y_{p} = - \frac{1}{4} x^{2} - \frac{3}{8} x - \frac{13}{32}$
	Answered by Bird last updated on 14/Oct/20
	$y^(′′) +3y^′ −4y =x^2 h→r^2 +3r−4=0 →Δ=9+16=25 ⇒r_1 =((−3+5)/2)=1 snd r_2 =((−3−5)/2)=−4 ⇒y_h =ae^x +b e^(−4x) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((e^x e^(−4x) )),((e^x −4e^(−4x) ))) =−4e^(−3x) −e^(−3x) =−5e^(−3x) ≠0 W_1 = determinant (((0 e^(−4x) )),((x^2 −4e^(−4x) )))=−x^2 e^(−4x) W_2 = determinant (((e^x 0)),((e^x x^2 )))=x^2 e^x v_1 =∫ (w_1 /w)dx =−∫ ((x^2 e^(−4x) )/(−5e^(−3x) ))dx =(1/5)∫x^2 e^(−x) dx =(1/5){−x^2 e^(−x) +∫ 2x e^(−x) } =(1/5){−x^2 e^(−x) +2(−xe^(−x) +∫e^(−x) dx)} =(1/5){−x^2 e^(−x) −2x e^(−x) −2e^(−x) } =(1/5)(−x^2 −2x−2)e^(−x) v_2 =∫ (w_2 /w)dx =∫ ((x^2 e^x )/(−5e^(−3x) ))dx =−(1/5)∫ x^2 e^(5x) dx =−(1/5){ (x^2 /5)e^(5x) −∫ ((2x)/5)e^(5x) dx} =−(1/5){ (x^2 /5)e^(5x) −(2/5)((x/5)e^(5x) −∫(1/5)e^(5x) dx)} =−(1/5){ ((x^2 e^(5x) )/5)−((2x)/(25))e^(5x) +(2/(125))e^(5x) } =−(1/5)((x^2 /5)−((2x)/(25))+(2/(125)))e^(5x) ⇒y_p =u_1 v_1 +u_2 v_2 =−(1/5)(x^2 +2x+2)−(1/5)((x^2 /5)−((2x)/(25))+(2/(25)))e^x general dolution is y =y_h +y_p$
	$y^{^{″}} + 3 y^{^{'}} - 4 y = x^{2}$ $h \to r^{2} + 3 r - 4 = 0 \to Δ = 9 + 16 = 25$ $\Rightarrow r_{1} = \frac{- 3 + 5}{2} = 1 s n d r_{2} = \frac{- 3 - 5}{2} = - 4$ $\Rightarrow y_{h} = {a e}^{x} + b e^{- 4 x} = {a u}_{1} + {b u}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{x} e^{- 4 x} \\ e^{x} - 4 e^{- 4 x} \end{matrix} \|$ $= - 4 e^{- 3 x} - e^{- 3 x} = - 5 e^{- 3 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 e^{- 4 x} \\ x^{2} - 4 e^{- 4 x} \end{matrix} \| = - x^{2} e^{- 4 x}$ $W_{2} = \| \begin{matrix} e^{x} 0 \\ e^{x} x^{2} \end{matrix} \| = x^{2} e^{x}$ $v_{1} = \int \frac{w_{1}}{w} d x = - \int \frac{x^{2} e^{- 4 x}}{- 5 e^{- 3 x}} d x$ $= \frac{1}{5} \int x^{2} e^{- x} d x$ $= \frac{1}{5} {- x^{2} e^{- x} + \int 2 x e^{- x}}$ $= \frac{1}{5} {- x^{2} e^{- x} + 2 (- {x e}^{- x} + \int e^{- x} d x)}$ $= \frac{1}{5} {- x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x}}$ $= \frac{1}{5} (- x^{2} - 2 x - 2) e^{- x}$ $v_{2} = \int \frac{w_{2}}{w} d x = \int \frac{x^{2} e^{x}}{- 5 e^{- 3 x}} d x$ $= - \frac{1}{5} \int x^{2} e^{5 x} d x$ $= - \frac{1}{5} {\frac{x^{2}}{5} e^{5 x} - \int \frac{2 x}{5} e^{5 x} d x}$ $= - \frac{1}{5} {\frac{x^{2}}{5} e^{5 x} - \frac{2}{5} (\frac{x}{5} e^{5 x} - \int \frac{1}{5} e^{5 x} d x)}$ $= - \frac{1}{5} {\frac{x^{2} e^{5 x}}{5} - \frac{2 x}{25} e^{5 x} + \frac{2}{125} e^{5 x}}$ $= - \frac{1}{5} (\frac{x^{2}}{5} - \frac{2 x}{25} + \frac{2}{125}) e^{5 x}$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $= - \frac{1}{5} (x^{2} + 2 x + 2) - \frac{1}{5} (\frac{x^{2}}{5} - \frac{2 x}{25} + \frac{2}{25}) e^{x}$ $g e n e r a l d o l u t i o n i s$ $y = y_{h} + y_{p}$
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