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Question Number 91848 by jagoll last updated on 03/May/20

y^(′′) −4y^′ +4y=(x+1)e^(2x)

$${y}^{''} −\mathrm{4}{y}^{'} +\mathrm{4}{y}=\left({x}+\mathrm{1}\right){e}^{\mathrm{2}{x}} \\ $$$$ \\ $$

Commented by john santu last updated on 03/May/20

complementary solution m^2 −4m+4=0 m_(1,2) = 2 ⇒y_c = A_1 e^(2x) +A_2 e^(2x) particular integral y_p = (B_1 x^3 +B_2 x^2 )e^(2x) y_p ^′ = (3B_1 x^2 +2B_2 x)e^(2x) +2(B_1 x^3 +B_2 x^2 )e^(2x) y_p ′′ = (6B_1 x+2B_2 )e^(2x) +2(3B_1 x^2 +2B_2 x)e^(2x) 2(3B_1 x^2 +2B_2 x)e^(2x) +4(B_1 x^3 +B_2 x^2 )e^(2x) comparing coeff B_2 = (1/2) ; B_1 =(1/6) ∴ y=(A_1 +A_2 )e^(2x) +((1/6)x^3 +(1/2)x^2 )e^(2x)

$${complementary}\:{solution} \\ $$$${m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{4}=\mathrm{0} \\ $$$${m}_{\mathrm{1},\mathrm{2}} \:=\:\mathrm{2}\:\Rightarrow{y}_{{c}} \:=\:{A}_{\mathrm{1}} {e}^{\mathrm{2}{x}} +{A}_{\mathrm{2}} {e}^{\mathrm{2}{x}} \\ $$$${particular}\:{integral} \\ $$$${y}_{{p}} =\:\left({B}_{\mathrm{1}} {x}^{\mathrm{3}} +{B}_{\mathrm{2}} {x}^{\mathrm{2}} \right){e}^{\mathrm{2}{x}} \\ $$$${y}_{{p}} ^{'} \:=\:\left(\mathrm{3}{B}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{2}{B}_{\mathrm{2}} {x}\right){e}^{\mathrm{2}{x}} \:+\mathrm{2}\left({B}_{\mathrm{1}} {x}^{\mathrm{3}} +{B}_{\mathrm{2}} {x}^{\mathrm{2}} \right){e}^{\mathrm{2}{x}} \\ $$$${y}_{{p}} ''\:=\:\left(\mathrm{6}{B}_{\mathrm{1}} {x}+\mathrm{2}{B}_{\mathrm{2}} \right){e}^{\mathrm{2}{x}} +\mathrm{2}\left(\mathrm{3}{B}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{2}{B}_{\mathrm{2}} {x}\right){e}^{\mathrm{2}{x}} \\ $$$$\mathrm{2}\left(\mathrm{3}{B}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{2}{B}_{\mathrm{2}} {x}\right){e}^{\mathrm{2}{x}} +\mathrm{4}\left({B}_{\mathrm{1}} {x}^{\mathrm{3}} +{B}_{\mathrm{2}} {x}^{\mathrm{2}} \right){e}^{\mathrm{2}{x}} \\ $$$${comparing}\:{coeff}\: \\ $$$${B}_{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:{B}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\therefore\:{y}=\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} \right){e}^{\mathrm{2}{x}} +\left(\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right){e}^{\mathrm{2}{x}} \\ $$

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