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Question Number 97626 by mathmax by abdo last updated on 08/Jun/20

solve y^(′′) −2y^′ +y =x^2 with y^′ (0) =y(0) =−1

$$\mathrm{solve}\:\mathrm{y}^{''} \:−\mathrm{2y}^{'} \:+\mathrm{y}\:\:=\mathrm{x}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\mathrm{y}\left(\mathrm{0}\right)\:=−\mathrm{1} \\ $$

Commented by bemath last updated on 09/Jun/20

λ^2 −2λ+1=0 λ=1,1 ⇒y_h = Ae^x + Bxe^x particular y_p =ax^2 +bx+c y′=2ax+b , y′′=2a ⇒2a−4ax−2b+ax^2 +bx+c = x^2 ⇒a=1 , b=4 , c = 2b−2a=6 y_p = x^2 +4x+6 generall solution y=Ae^x +Bxe^x +x^2 +4x+6 y(0) = −1=A+6 , A = −7 y′= −7e^x +Be^x +Bxe^x +2x+4 y′(0)=−1=−7+B+4 B=2 ⇔∴y_c = −7e^x +2xe^x +x^2 +4x+6

$$\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{1}=\mathrm{0} \\ $$$$\lambda=\mathrm{1},\mathrm{1}\:\Rightarrow\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{Ae}^{\mathrm{x}} \:+\:\mathrm{Bxe}^{\mathrm{x}} \\ $$$$\mathrm{particular}\:\mathrm{y}_{\mathrm{p}} =\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c} \\ $$$$\mathrm{y}'=\mathrm{2ax}+\mathrm{b}\:,\:\mathrm{y}''=\mathrm{2a} \\ $$$$\Rightarrow\mathrm{2a}−\mathrm{4ax}−\mathrm{2b}+\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1}\:,\:\mathrm{b}=\mathrm{4}\:,\:\mathrm{c}\:=\:\mathrm{2b}−\mathrm{2a}=\mathrm{6} \\ $$$$\mathrm{y}_{\mathrm{p}} =\:\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{6}\: \\ $$$$\mathrm{generall}\:\mathrm{solution}\: \\ $$$$\mathrm{y}=\mathrm{Ae}^{\mathrm{x}} +\mathrm{Bxe}^{\mathrm{x}} +\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{6} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:−\mathrm{1}=\mathrm{A}+\mathrm{6}\:,\:\mathrm{A}\:=\:−\mathrm{7} \\ $$$$\mathrm{y}'=\:−\mathrm{7e}^{\mathrm{x}} +\mathrm{Be}^{\mathrm{x}} +\mathrm{Bxe}^{\mathrm{x}} +\mathrm{2x}+\mathrm{4} \\ $$$$\mathrm{y}'\left(\mathrm{0}\right)=−\mathrm{1}=−\mathrm{7}+\mathrm{B}+\mathrm{4} \\ $$$$\mathrm{B}=\mathrm{2}\:\Leftrightarrow\therefore\mathrm{y}_{\mathrm{c}} \:=\:−\mathrm{7e}^{\mathrm{x}} +\mathrm{2xe}^{\mathrm{x}} +\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{6} \\ $$

Commented by bobhans last updated on 09/Jun/20

good

$$\mathrm{good} \\ $$

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