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


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Question Number 105878 by bemath last updated on 01/Aug/20

$\frac{d^{3} y}{{d x}^{3}} + 3 \frac{d^{2} y}{{d x}^{2}} + 2 \frac{d y}{d x} = x^{2}$
	Answered by bobhans last updated on 01/Aug/20
	$HE : ♭^3 +3♭^2 +2♭ = 0 ♭ (♭^2 +3♭+2) = 0 → { ((♭=0)),((♭=−1)),((♭=−2)) :} Y_h = A+Be^(−x) +Ce^(−2x) Particular solution. by guessing Y_p = px^3 +qx^2 +rx Y_p ′=3px^2 +2qx+r Y_p ′′ = 6px+2q ; Y_p ′′′= 6p comparing coefficient 6p+3(6px+2q)+2(3px^2 +2qx+r) = x^2 6px^2 +(18p+4q)x+6p+6q+2r = x^2 { ((6p = 1 →p=(1/6))),((18p+4q=0→q=−((9.((1/6)))/2)=−(3/4))),((r=−3p−3q=−(1/2)+(9/4)=(7/4))) :} Y_p = (x^3 /6)−((3x^2 )/4)+((7x)/4) General solution Y_g = A+Be^(−x) +Ce^(−2x) +(x^3 /6)−((3x^2 )/4)+((7x)/4) ⊸$
	$HE : ♭^{3} + 3 ♭^{2} + 2 ♭ = 0$ $♭ (♭^{2} + 3 ♭ + 2) = 0 \to {\begin{cases} ♭ = 0 \\ ♭ = - 1 \\ ♭ = - 2 \end{cases}$ $Y_{h} = A + {Be}^{- x} + {Ce}^{- 2 x}$ $Particular solution . by guessing$ $Y_{p} = {px}^{3} + {qx}^{2} + rx$ $Y_{p}^{'} = {3 px}^{2} + 2 qx + r$ $Y_{p}^{″} = 6 px + 2 q; Y_{p}^{‴} = 6 p$ $comparing coefficient$ $6 p + 3 (6 px + 2 q) + 2 ({3 px}^{2} + 2 qx + r) = x^{2}$ ${6 px}^{2} + (18 p + 4 q) x + 6 p + 6 q + 2 r = x^{2}$ ${\begin{cases} 6 p = 1 \to p = \frac{1}{6} \\ 18 p + 4 q = 0 \to q = - \frac{9 . (\frac{1}{6})}{2} = - \frac{3}{4} \\ r = - 3 p - 3 q = - \frac{1}{2} + \frac{9}{4} = \frac{7}{4} \end{cases}$ $Y_{p} = \frac{x^{3}}{6} - \frac{{3 x}^{2}}{4} + \frac{7 x}{4}$ $G eneral solution$ $Y_{g} = A + {Be}^{- x} + {Ce}^{- 2 x} + \frac{x^{3}}{6} - \frac{{3 x}^{2}}{4} + \frac{7 x}{4} ⊸$
	Answered by mathmax by abdo last updated on 02/Aug/20
	$y^((3)) +3y^((2)) +2y^((1)) =x^2 let y^′ =z ⇒z^(′′) +3z^′ +2z =x^2 h→r^2 +3r+2 =0→Δ =9−8=1 ⇒r_1 =((−3+1)/2) =−1 r_2 =((−3−1)/2) =−2 ⇒z_h =ae^(−x) +be^(−2x) =az_1 +bz_2 W(z_1 ,z_2 ) = determinant (((e^(−x) e^(−2x) )),((−e^(−x) −2e^(−2x) )))=−2e^(−3x) +e^(−3x) =−e^(−3x) ≠0 W_1 = determinant (((0 e^(−2x) )),((x^2 −2e^(−2x) )))=−x^2 e^(−2x) W_2 = determinant (((e^(−x) 0)),((−e^(−x) x^2 )))=x^2 e^(−x) v_1 =∫ (w_1 /w) dx =−∫ ((x^2 e^(−2x) )/(−e^(−3x) ))dx =∫ x^2 e^x dx =x^2 e^x −∫ 2x e^x dx =x^2 e^x −2 { xe^x −∫ e^x dx} =(x^2 −2x+2)e^x v_2 =∫ (w_2 /w)dx =∫ ((x^2 e^(−x) )/(−e^(−3x) ))dx =−∫ x^2 e^(2x) =−{ (x^2 /2)e^(2x) −∫ 2x×(1/2)e^(2x) dx} =−{(x^2 /2)e^(2x) −{ (x/2)e^(2x) −∫ (1/2)e^(2x) dx}} =−{(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) } ={−(x^2 /2)+(x/2)−(1/4)}e^(2x) z_p =z_1 v_(1 ) +z_2 v_2 =e^(−x) (x^2 −2x+2)e^x +e^(−2x) (−(x^2 /2)+(x/2)−(1/4))e^(2x) =x^2 −2x+2 −(x^2 /2) +(x/2)−(1/4) =(x^2 /2)−(3/2)x +(7/4) ⇒z =z_x +z_p =ae^(−x) +be^(−2x) +(x^2 /2)−((3x)/2) +(7/4) =y^′ ⇒ y =∫ z(x)dx =−a e^(−x) −(b/2)e^(−2x) +(x^3 /6)−(3/4)x^2 +(7/4)x +C$
	$y^{(3)} + {3 y}^{(2)} + {2 y}^{(1)} = x^{2} let y^{^{'}} = z \Rightarrow z^{^{″}} + {3 z}^{^{'}} + 2 z = x^{2}$ $h \to r^{2} + 3 r + 2 = 0 \to Δ = 9 - 8 = 1 \Rightarrow r_{1} = \frac{- 3 + 1}{2} = - 1$ $r_{2} = \frac{- 3 - 1}{2} = - 2 \Rightarrow z_{h} = {ae}^{- x} + {be}^{- 2 x} = {az}_{1} + {bz}_{2}$ $W (z_{1}, z_{2}) = \| \begin{matrix} e^{- x} e^{- 2 x} \\ - e^{- x} - {2 e}^{- 2 x} \end{matrix} \| = - {2 e}^{- 3 x} + e^{- 3 x} = - e^{- 3 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 e^{- 2 x} \\ x^{2} - {2 e}^{- 2 x} \end{matrix} \| = - x^{2} e^{- 2 x}$ $W_{2} = \| \begin{matrix} e^{- x} 0 \\ - e^{- x} x^{2} \end{matrix} \| = x^{2} e^{- x}$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{x^{2} e^{- 2 x}}{- e^{- 3 x}} dx = \int x^{2} e^{x} dx$ $= x^{2} e^{x} - \int 2 x e^{x} dx = x^{2} e^{x} - 2 {{xe}^{x} - \int e^{x} dx}$ $= (x^{2} - 2 x + 2) e^{x}$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{x^{2} e^{- x}}{- e^{- 3 x}} dx = - \int x^{2} e^{2 x}$ $= - {\frac{x^{2}}{2} e^{2 x} - \int 2 x \times \frac{1}{2} e^{2 x} dx} = - {\frac{x^{2}}{2} e^{2 x} - {\frac{x}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} dx}}$ $= - {\frac{x^{2}}{2} e^{2 x} - \frac{x}{2} e^{2 x} + \frac{1}{4} e^{2 x}} = {- \frac{x^{2}}{2} + \frac{x}{2} - \frac{1}{4}} e^{2 x}$ $z_{p} = z_{1} v_{1} + z_{2} v_{2} = e^{- x} (x^{2} - 2 x + 2) e^{x} + e^{- 2 x} (- \frac{x^{2}}{2} + \frac{x}{2} - \frac{1}{4}) e^{2 x}$ $= x^{2} - 2 x + 2 - \frac{x^{2}}{2} + \frac{x}{2} - \frac{1}{4} = \frac{x^{2}}{2} - \frac{3}{2} x + \frac{7}{4}$ $\Rightarrow z = z_{x} + z_{p} = {ae}^{- x} + {be}^{- 2 x} + \frac{x^{2}}{2} - \frac{3 x}{2} + \frac{7}{4} = y^{^{'}} \Rightarrow$ $y = \int z (x) dx = - a e^{- x} - \frac{b}{2} e^{- 2 x} + \frac{x^{3}}{6} - \frac{3}{4} x^{2} + \frac{7}{4} x + C$
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