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Question Number 90983 by jagoll last updated on 27/Apr/20

y′′−5y′+6y=x^2

$${y}''−\mathrm{5}{y}'+\mathrm{6}{y}={x}^{\mathrm{2}} \\ $$

Commented by john santu last updated on 27/Apr/20

complementari solution y = Ae^(q_1 x) + Be^(q_2 x) ⇒q^2 −5q +6 = 0 q = { (3),(2) :} ⇒y = Ae^(3x) + Be^(2x) particular solution y_p = ax^2 +bx +c y′= 2ax+ b , y′′= 2a 2a−5(2ax+b)+6ax^2 +6bx+6c = x^2 a = (1/6), b = (5/(18)) , c = ((19)/(108)) general solution y=Ae^(3x) + Be^(2x) + (1/6)x^2 +(5/(18))x+((19)/(108))

$$\mathrm{complementari}\:\mathrm{solution} \\ $$$$\mathrm{y}\:=\:\mathrm{Ae}^{\mathrm{q}_{\mathrm{1}} \mathrm{x}} \:+\:\mathrm{Be}^{\mathrm{q}_{\mathrm{2}} \mathrm{x}} \\ $$$$\Rightarrow\mathrm{q}^{\mathrm{2}} −\mathrm{5q}\:+\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{q}\:=\:\begin{cases}{\mathrm{3}}\\{\mathrm{2}}\end{cases}\:\Rightarrow\mathrm{y}\:=\:\mathrm{Ae}^{\mathrm{3x}} \:+\:\mathrm{Be}^{\mathrm{2x}} \\ $$$$\mathrm{particular}\:\mathrm{solution}\: \\ $$$$\mathrm{y}_{\mathrm{p}} =\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}\:+\mathrm{c}\: \\ $$$$\mathrm{y}'=\:\mathrm{2ax}+\:\mathrm{b}\:,\:\mathrm{y}''=\:\mathrm{2a} \\ $$$$\mathrm{2a}−\mathrm{5}\left(\mathrm{2ax}+\mathrm{b}\right)+\mathrm{6ax}^{\mathrm{2}} +\mathrm{6bx}+\mathrm{6c}\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{6}},\:\mathrm{b}\:=\:\frac{\mathrm{5}}{\mathrm{18}}\:,\:\mathrm{c}\:=\:\frac{\mathrm{19}}{\mathrm{108}} \\ $$$$\mathrm{general}\:\mathrm{solution} \\ $$$$\mathrm{y}=\mathrm{Ae}^{\mathrm{3x}} \:+\:\mathrm{Be}^{\mathrm{2x}} +\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{18}}\mathrm{x}+\frac{\mathrm{19}}{\mathrm{108}} \\ $$

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