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Question Number 129187 by math178 last updated on 13/Jan/21

Commented by math178 last updated on 13/Jan/21

$d i f f e r a n t i a l$ $w h a t i s t h e s p e c i a l s o l u t i o n ?$ $t h a n k y o u < 3$
	Answered by Olaf last updated on 13/Jan/21
	$y^((3)) −4y′′+y′+6y = x^3 −4x+2 (1) y_0 = ax^3 +bx^2 +cx+d y_0 ′ = 3ax^2 +2bx+c y_0 ′′ = 6ax+2b y_0 ^((3)) = 6a y_0 is solution of equation (1) ⇔ 6a−4(6ax+2b)+(3ax^2 +2bx+c)+ +6(ax^3 +bx^2 +cx+d) = x^3 −4x+2 { ((6a = 1 (2))),((3a+6b = 0 (3))),((−24a+2b+6c = −4 (4))),((6a−8b+c+6d = 2 (5))) :} (2) : a = (1/6) (3) : b = −(1/(12)) (4) : c = (1/(36)) (5) : d = ((11)/(216)) y_0 = (x^3 /6)−(x^2 /(12))+(x/(36))+((11)/(216)) y_H ^((3)) −4y_H ′′+y_H ′+6y_H = 0 (6) We solve r^3 −4r^2 +r+6 = 0 (7) (r+1)(r^2 −5r+6) = 0 (r+1)(r−2)(r−3) = 0 ⇒ the solutions of (7) are −1, 2 and 3 and y_H = Ae^(−x) +Be^(2x) +Ce^(3x) y = y_H +y_0 = Ae^(−x) +Be^(2x) +Ce^(3x) +(x^3 /6)−(x^2 /(12))+(x/(36))+((11)/(216))$
	$y^{(3)} - 4 y^{″} + y^{'} + 6 y = x^{3} - 4 x + 2 (1)$ $y_{0} = {a x}^{3} + {b x}^{2} + c x + d$ $y_{0}^{'} = 3 {a x}^{2} + 2 b x + c$ $y_{0}^{″} = 6 a x + 2 b$ $y_{0}^{(3)} = 6 a$ $y_{0} is solution of equation (1)$ $\Leftrightarrow 6 a - 4 (6 a x + 2 b) + (3 {a x}^{2} + 2 b x + c) +$ $+ 6 ({a x}^{3} + {b x}^{2} + c x + d) = x^{3} - 4 x + 2$ ${\begin{cases} 6 a = 1 (2) \\ 3 a + 6 b = 0 (3) \\ - 24 a + 2 b + 6 c = - 4 (4) \\ 6 a - 8 b + c + 6 d = 2 (5) \end{cases}$ $(2) : a = \frac{1}{6}$ $(3) : b = - \frac{1}{12}$ $(4) : c = \frac{1}{36}$ $(5) : d = \frac{11}{216}$ $y_{0} = \frac{x^{3}}{6} - \frac{x^{2}}{12} + \frac{x}{36} + \frac{11}{216}$ $y_{H}^{(3)} - 4 y_{H}^{″} + y_{H}^{'} + 6 y_{H} = 0 (6)$ $We solve r^{3} - 4 r^{2} + r + 6 = 0 (7)$ $(r + 1) (r^{2} - 5 r + 6) = 0$ $(r + 1) (r - 2) (r - 3) = 0$ $\Rightarrow the solutions of (7) are - 1, 2 and 3$ $and y_{H} = A e^{- x} + B e^{2 x} + C e^{3 x}$ $y = y_{H} + y_{0} = A e^{- x} + B e^{2 x} + C e^{3 x} + \frac{x^{3}}{6} - \frac{x^{2}}{12} + \frac{x}{36} + \frac{11}{216}$
	Commented by math178 last updated on 13/Jan/21

	$t h a n k y o u =)$
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