y′′+y′−6y = x


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Question Number 105879 by john santu last updated on 01/Aug/20

$y^{″} + y^{'} - 6 y = x$
	Answered by bemath last updated on 01/Aug/20

	Answered by mathmax by abdo last updated on 01/Aug/20
	$y^(′′) +y^′ −6y =x h→y^(′′) +y^′ −6y =0→r^2 +r−6 =0→Δ =1+24=25 ⇒ r_1 =((−1+5)/2) =2 and r_2 =((−1−5)/2) =−3 ⇒y_h =ae^(2x) +be^(−3x) =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((e^(2x) e^(−3x) )),((2e^(2x) −3e^(−3x) )))=−3e^(−x) −2e^(−x) =−5e^(−x) ≠0 W_1 = determinant (((o e^(−3x) )),((x −3e^(−3x) )))=−xe^(−3x) W_2 = determinant (((e^(2x) 0)),((2e^(2x) x)))=xe^(2x) ⇒ v_1 =∫ (w_1 /w)dx =∫ ((−xe^(−3x) )/(−5e^(−x) ))dx =(1/5)∫x e^(−2x) dx ⇒5v_1 =−(x/2)e^(−2x) +∫ (1/2)e^(−2x) dx =−(x/2)e^(−2x) −(1/4)e^(−2x) =−((x/2)+(1/4))e^(−2x) ⇒v_1 =−((x/(10)) +(1/(2o)))e^(−2x) v_2 =∫ (w_2 /w)dx =∫ ((xe^(2x) )/(−5e^(−x) ))dx =−(1/5) ∫ xe^(3x) dx =−(1/5){ (x/3)e^(3x) −∫ (1/3)e^(3x) dx} =−(1/5){(x/3)e^(3x) −(1/9)e^(3x) } ={(1/(45))−(x/(15))}e^(3x) ⇒y_p =u_1 v_(1 ) +u_2 v_2 =−e^(2x) ×((x/(10)) +(1/(20)))e^(−2x) + e^(−3x) ×{(1/(45))−(x/(15))}e^(3x) =−(x/(10))−(1/(20)) +(1/(45))−(x/(15)) =−((1/(10))+(1/(15)))x +(1/(45))−(1/(20)) =−((25)/(150)) x −((25)/(900))(rest simplification!) the general solution is y =ae^(2x) +b e^(−3x) −((25)/(150))x−((25)/(900))$
	$y^{^{″}} + y^{^{'}} - 6 y = x$ $h \to y^{^{″}} + y^{^{'}} - 6 y = 0 \to r^{2} + r - 6 = 0 \to Δ = 1 + 24 = 25 \Rightarrow$ $r_{1} = \frac{- 1 + 5}{2} = 2 and r_{2} = \frac{- 1 - 5}{2} = - 3 \Rightarrow y_{h} = {ae}^{2 x} + {be}^{- 3 x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{2 x} e^{- 3 x} \\ {2 e}^{2 x} - {3 e}^{- 3 x} \end{matrix} \| = - {3 e}^{- x} - {2 e}^{- x} = - {5 e}^{- x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{- 3 x} \\ x - {3 e}^{- 3 x} \end{matrix} \| = - {xe}^{- 3 x}$ $W_{2} = \| \begin{matrix} e^{2 x} 0 \\ {2 e}^{2 x} x \end{matrix} \| = {xe}^{2 x} \Rightarrow$ $v_{1} = \int \frac{w_{1}}{w} dx = \int \frac{- {xe}^{- 3 x}}{- {5 e}^{- x}} dx = \frac{1}{5} \int x e^{- 2 x} dx$ $\Rightarrow {5 v}_{1} = - \frac{x}{2} e^{- 2 x} + \int \frac{1}{2} e^{- 2 x} dx = - \frac{x}{2} e^{- 2 x} - \frac{1}{4} e^{- 2 x}$ $= - (\frac{x}{2} + \frac{1}{4}) e^{- 2 x} \Rightarrow v_{1} = - (\frac{x}{10} + \frac{1}{2 o}) e^{- 2 x}$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{{xe}^{2 x}}{- {5 e}^{- x}} dx = - \frac{1}{5} \int {xe}^{3 x} dx$ $= - \frac{1}{5} {\frac{x}{3} e^{3 x} - \int \frac{1}{3} e^{3 x} dx} = - \frac{1}{5} {\frac{x}{3} e^{3 x} - \frac{1}{9} e^{3 x}}$ $= {\frac{1}{45} - \frac{x}{15}} e^{3 x} \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $= - e^{2 x} \times (\frac{x}{10} + \frac{1}{20}) e^{- 2 x} + e^{- 3 x} \times {\frac{1}{45} - \frac{x}{15}} e^{3 x}$ $= - \frac{x}{10} - \frac{1}{20} + \frac{1}{45} - \frac{x}{15} = - (\frac{1}{10} + \frac{1}{15}) x + \frac{1}{45} - \frac{1}{20}$ $= - \frac{25}{150} x - \frac{25}{900} (rest simplification!)$ $the general solution is y = {ae}^{2 x} + b e^{- 3 x} - \frac{25}{150} x - \frac{25}{900}$
	Commented by Coronavirus last updated on 01/Aug/20
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	Answered by john santu last updated on 01/Aug/20
	$particular solution y_p = Ax^2 +Bx+C y′=2Ax+B ⇒y′′=2A comparing coefficient 2A+2Ax+B−6Ax^2 −6Bx−6C=x −6Ax^2 +(2A−6B)x+2A+B−6C=x { ((A=0)),((2A−6B=1→B=−(1/6))),((2A+B−6C=0→C=−(1/(36)))) :} y_p = −(1/6)x−(1/(36)) ■$
	$p a r t i c u l a r s o l u t i o n$ $y_{p} = {A x}^{2} + B x + C$ $y^{'} = 2 A x + B \Rightarrow y^{″} = 2 A$ $c o m p a r i n g c o e f f i c i e n t$ $2 A + 2 A x + B - 6 {A x}^{2} - 6 B x - 6 C = x$ $- 6 {A x}^{2} + (2 A - 6 B) x + 2 A + B - 6 C = x$ ${\begin{cases} A = 0 \\ 2 A - 6 B = 1 \to B = - \frac{1}{6} \\ 2 A + B - 6 C = 0 \to C = - \frac{1}{36} \end{cases}$ $y_{p} = - \frac{1}{6} x - \frac{1}{36} ◼$
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