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find-in-the-form-y-f-x-the-general-solution-of-the-differentail-equation-d-2-y-dx-2-dy-dx-6y-e-3x-

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Question Number 87751 by Rio Michael last updated on 06/Apr/20
find in the form y= f(x) the general solution of the differentail equation (d^2 y/dx^2 ) −(dy/dx)−6y = e^(3x)
$$\mathrm{find}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{y}=\:{f}\left({x}\right)\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{differentail}\:\mathrm{equation} \\ $$$$\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\frac{{dy}}{{dx}}−\mathrm{6}{y}\:=\:{e}^{\mathrm{3}{x}} \\ $$$$ \\ $$
Commented by niroj last updated on 06/Apr/20
(d^2 y/dx^2 )−(dy/dx)−6y=e^(3x) let, (d^2 y/dx^2 )=D^2 y and (dy/dx)=Dy (D^2 −D−6)y=e^(3x) Auxilairy Equation, m^2 −m−6=0 m^2 −3m+2m−6=0 m(m−3)+2(m−3)=0 (m−3)(m+2)=0 m= 3,−2 Complementary Fuction = C_1 e^(3x) +C_2 e^(−2x) Particular Integral= (e^(3x) /(D^2 −D−6)) = ((x.e^(3x) )/(2D−1))= ((xe^(3x) )/(6−1))= ((xe^(3x) )/5) y= f(x)=CF+PI y= C_1 e^(3x) +C_2 e^(−2x) + ((xe^(3x) )/5)//.
$$\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }−\frac{\mathrm{dy}}{\mathrm{dx}}−\mathrm{6y}=\mathrm{e}^{\mathrm{3x}} \\ $$$$\:\:\mathrm{let},\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\mathrm{D}^{\mathrm{2}} \mathrm{y}\:\mathrm{and}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{Dy} \\ $$$$\:\:\:\left(\mathrm{D}^{\mathrm{2}} −\mathrm{D}−\mathrm{6}\right)\mathrm{y}=\mathrm{e}^{\mathrm{3x}} \\ $$$$\:\:\mathrm{Auxilairy}\:\mathrm{Equation}, \\ $$$$\:\:\:\:\:\mathrm{m}^{\mathrm{2}} −\mathrm{m}−\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{m}^{\mathrm{2}} −\mathrm{3m}+\mathrm{2m}−\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\mathrm{m}\left(\mathrm{m}−\mathrm{3}\right)+\mathrm{2}\left(\mathrm{m}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{m}−\mathrm{3}\right)\left(\mathrm{m}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{m}=\:\mathrm{3},−\mathrm{2} \\ $$$$\:\:\mathrm{Complementary}\:\mathrm{Fuction}\:=\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{\mathrm{3x}} +\mathrm{C}_{\mathrm{2}} \mathrm{e}^{−\mathrm{2x}} \\ $$$$\:\:\mathrm{Particular}\:\mathrm{Integral}=\:\frac{\mathrm{e}^{\mathrm{3x}} }{\mathrm{D}^{\mathrm{2}} −\mathrm{D}−\mathrm{6}} \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{x}.\mathrm{e}^{\mathrm{3x}} }{\mathrm{2D}−\mathrm{1}}=\:\frac{\mathrm{xe}^{\mathrm{3x}} }{\mathrm{6}−\mathrm{1}}=\:\frac{\mathrm{xe}^{\mathrm{3x}} }{\mathrm{5}} \\ $$$$\:\:\:\:\mathrm{y}=\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{CF}+\mathrm{PI} \\ $$$$\:\:\:\:\mathrm{y}=\:\mathrm{C}_{\mathrm{1}} \mathrm{e}^{\mathrm{3x}} +\mathrm{C}_{\mathrm{2}} \mathrm{e}^{−\mathrm{2x}} +\:\frac{\mathrm{xe}^{\mathrm{3x}} }{\mathrm{5}}//. \\ $$
Commented by Rio Michael last updated on 06/Apr/20
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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