Question Number 93815 by i jagooll last updated on 15/May/20
$$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{6D}+\mathrm{9}\right)\mathrm{y}\:=\:\frac{\mathrm{e}^{−\mathrm{3x}} }{\mathrm{x}^{\mathrm{3}} }\: \\ $$
Answered by john santu last updated on 15/May/20
![homogenous solution λ^2 +6λ+9=0 λ = −3,−3 y_h = Ae^(−3x) +Bxe^(−3x ) particular solution (D+3)[ (D+3)(e^(3x) )]=(1/x^3 ) (D+3)[D(e^(3x) y)] = (1/x^3 ) (D+3)(e^(3x) y)=∫ (1/x^3 ) dx=−(1/(2x^2 )) e^(3x) y = ∫ −(1/(2x^2 )) dx = y_p = (1/(2x.e^(3x) )) = (e^(−3x) /(2x)) generall solution y = Ae^(−3x) +Bxe^(−3x) +(e^(−3x) /(2x))](https://www.tinkutara.com/question/Q93816.png)
$$\mathrm{homogenous}\:\mathrm{solution} \\ $$$$\lambda^{\mathrm{2}} +\mathrm{6}\lambda+\mathrm{9}=\mathrm{0} \\ $$$$\lambda\:=\:−\mathrm{3},−\mathrm{3}\: \\ $$$$\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{Ae}^{−\mathrm{3x}} +\mathrm{Bxe}^{−\mathrm{3x}\:} \\ $$$$\mathrm{particular}\:\mathrm{solution} \\ $$$$\left(\mathrm{D}+\mathrm{3}\right)\left[\:\left(\mathrm{D}+\mathrm{3}\right)\left(\mathrm{e}^{\mathrm{3x}} \right)\right]=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\left(\mathrm{D}+\mathrm{3}\right)\left[\mathrm{D}\left(\mathrm{e}^{\mathrm{3x}} \mathrm{y}\right)\right]\:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\: \\ $$$$\left(\mathrm{D}+\mathrm{3}\right)\left(\mathrm{e}^{\mathrm{3x}} \mathrm{y}\right)=\int\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} } \\ $$$$\mathrm{e}^{\mathrm{3x}} \mathrm{y}\:=\:\int\:−\frac{\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }\:\mathrm{dx}\:= \\ $$$$\mathrm{y}_{\mathrm{p}} \:=\:\frac{\mathrm{1}}{\mathrm{2x}.\mathrm{e}^{\mathrm{3x}} }\:=\:\frac{\mathrm{e}^{−\mathrm{3x}} }{\mathrm{2x}} \\ $$$$\mathrm{generall}\:\mathrm{solution} \\ $$$$\mathrm{y}\:=\:\mathrm{Ae}^{−\mathrm{3x}} +\mathrm{Bxe}^{−\mathrm{3x}} +\frac{\mathrm{e}^{−\mathrm{3x}} }{\mathrm{2x}} \\ $$
Commented by i jagooll last updated on 15/May/20
$$\mathrm{thAnk}\:\mathrm{you} \\ $$