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Question Number 115721 by bemath last updated on 28/Sep/20
(D^2 −6D+9)y = (e^(3x) /x^2 )
$$\left({D}^{\mathrm{2}} −\mathrm{6}{D}+\mathrm{9}\right){y}\:=\:\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} } \\ $$
Commented by mohammad17 last updated on 28/Sep/20
m^2 −6m+9=0⇒(m−3)^2 =0⇒m_1 =m_2 =3 Yc=c_1 e^(3x) +c_2 xe^(3x) W(u_1 ,u_2 )= determinant (((e^(3x) xe^(3x) )),((3e^(3x) 3xe^(3x) +e^(3x) )))=e^(6x) ≠0 W_1 = determinant (((0 xe^(3x) )),(((e^(3x) /x^2 ) 3xe^(3x) +e^(3x) )))=−(e^(6x) /x) W_2 = determinant (((e^(3x) 0)),((3e^(3x) (e^(3x) /x^2 ))))=(e^(6x) /x^2 ) V_1 =∫ (W_1 /W) dx=∫ −(1/x^2 )dx=(1/x) V_2 =∫ (W_2 /W) dx=∫(1/x^2 )dx=−(1/x) Yp=V_1 U_1 +V_2 U_2 Yp=(e^(3x) /x)−e^(3x) Y=Yc+Yp Y=c_1 e^(3x) +c_2 xe^(3x) +(e^(3x) /x)−e^(3x) (m.o)
$${m}^{\mathrm{2}} −\mathrm{6}{m}+\mathrm{9}=\mathrm{0}\Rightarrow\left({m}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{m}_{\mathrm{1}} ={m}_{\mathrm{2}} =\mathrm{3} \\ $$$${Yc}={c}_{\mathrm{1}} {e}^{\mathrm{3}{x}} +{c}_{\mathrm{2}} {xe}^{\mathrm{3}{x}} \\ $$$$ \\ $$$${W}\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right)=\begin{vmatrix}{{e}^{\mathrm{3}{x}} \:\:\:\:\:{xe}^{\mathrm{3}{x}} }\\{\mathrm{3}{e}^{\mathrm{3}{x}} \:\:\:\:\mathrm{3}{xe}^{\mathrm{3}{x}} +{e}^{\mathrm{3}{x}} }\end{vmatrix}={e}^{\mathrm{6}{x}} \neq\mathrm{0} \\ $$$$ \\ $$$${W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{xe}^{\mathrm{3}{x}} }\\{\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\mathrm{3}{xe}^{\mathrm{3}{x}} +{e}^{\mathrm{3}{x}} }\end{vmatrix}=−\frac{{e}^{\mathrm{6}{x}} }{{x}} \\ $$$$ \\ $$$${W}_{\mathrm{2}} =\begin{vmatrix}{{e}^{\mathrm{3}{x}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{3}{e}^{\mathrm{3}{x}} \:\:\:\:\:\:\:\:\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} }}\end{vmatrix}=\frac{{e}^{\mathrm{6}{x}} }{{x}^{\mathrm{2}} } \\ $$$$ \\ $$$${V}_{\mathrm{1}} =\int\:\frac{{W}_{\mathrm{1}} }{{W}}\:{dx}=\int\:−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{{x}} \\ $$$$ \\ $$$${V}_{\mathrm{2}} =\int\:\frac{{W}_{\mathrm{2}} }{{W}}\:{dx}=\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{{x}} \\ $$$$ \\ $$$${Yp}={V}_{\mathrm{1}} {U}_{\mathrm{1}} +{V}_{\mathrm{2}} {U}_{\mathrm{2}} \\ $$$$ \\ $$$${Yp}=\frac{{e}^{\mathrm{3}{x}} }{{x}}−{e}^{\mathrm{3}{x}} \\ $$$$ \\ $$$${Y}={Yc}+{Yp} \\ $$$$ \\ $$$${Y}={c}_{\mathrm{1}} {e}^{\mathrm{3}{x}} +{c}_{\mathrm{2}} {xe}^{\mathrm{3}{x}} +\frac{{e}^{\mathrm{3}{x}} }{{x}}−{e}^{\mathrm{3}{x}} \\ $$$$ \\ $$$$\left({m}.{o}\right) \\ $$
Commented by mohammad17 last updated on 28/Sep/20
Yc=c_1 e^(3x) +c_2 xe^(3x) Yp=e^(ax) . (1/(f(D+a))) f(x) Yp=e^(3x) (1/((D+3)^2 −6(D+3)+9)).((1/x^2 )) Yp=e^(3x) (1/D^2 )((1/x^2 ))⇒Yp=−e^(3x) (1/D)((1/x)) Yp=−e^(3x) lnx Y=Yc+Yp Y=c_1 e^(3x) +c_2 xe^(3x) −e^(3x) lnx
$${Yc}={c}_{\mathrm{1}} {e}^{\mathrm{3}{x}} +{c}_{\mathrm{2}} {xe}^{\mathrm{3}{x}} \\ $$$$ \\ $$$${Yp}={e}^{{ax}} \:.\:\frac{\mathrm{1}}{{f}\left({D}+{a}\right)}\:{f}\left({x}\right) \\ $$$$ \\ $$$${Yp}={e}^{\mathrm{3}{x}} \frac{\mathrm{1}}{\left({D}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{6}\left({D}+\mathrm{3}\right)+\mathrm{9}}.\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${Yp}={e}^{\mathrm{3}{x}} \:\frac{\mathrm{1}}{{D}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\Rightarrow{Yp}=−{e}^{\mathrm{3}{x}} \frac{\mathrm{1}}{{D}}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$ \\ $$$${Yp}=−{e}^{\mathrm{3}{x}} {lnx} \\ $$$$ \\ $$$${Y}={Yc}+{Yp} \\ $$$$ \\ $$$${Y}={c}_{\mathrm{1}} {e}^{\mathrm{3}{x}} +{c}_{\mathrm{2}} {xe}^{\mathrm{3}{x}} −{e}^{\mathrm{3}{x}} {lnx} \\ $$
Commented by bemath last updated on 28/Sep/20
gave kudos all master
$${gave}\:{kudos}\:{all}\:{master} \\ $$
Commented by mohammad17 last updated on 28/Sep/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by john santu last updated on 28/Sep/20
let y = e^(3x) .q → { ((Dy=e^(3x) (q′+3q))),((D^2 y=e^(3x) (q′′+6q′+9q))) :} we obtain e^(3x) (q′′+6q′+9q)−6.e^(3x) (q′+3q)+9e^(3x) q=(e^(3x) /x^2 ) which simplifies to q′′ = (1/x^2 ) → ((d(dq))/dx) = (1/x^2 ) ∫ ((d(dq))/dx) = ∫ (1/x^2 ) ⇒ (dq/dx) = −(1/x) +C_1 ∫ dq = ∫ (−(1/x) +C_1 ) dx →q = −ln ∣x∣ + C_1 x+C_2 ⇒(y/e^(3x) ) = −ln ∣x∣ + C_1 x+C_2 ⇒y = −e^(3x) .ln ∣x∣ + (C_1 x+C_2 ).e^(3x)
$${let}\:{y}\:=\:{e}^{\mathrm{3}{x}} .{q}\:\rightarrow\begin{cases}{{Dy}={e}^{\mathrm{3}{x}} \left({q}'+\mathrm{3}{q}\right)}\\{{D}^{\mathrm{2}} {y}={e}^{\mathrm{3}{x}} \left({q}''+\mathrm{6}{q}'+\mathrm{9}{q}\right)}\end{cases} \\ $$$${we}\:{obtain}\: \\ $$$${e}^{\mathrm{3}{x}} \left({q}''+\mathrm{6}{q}'+\mathrm{9}{q}\right)−\mathrm{6}.{e}^{\mathrm{3}{x}} \left({q}'+\mathrm{3}{q}\right)+\mathrm{9}{e}^{\mathrm{3}{x}} {q}=\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} } \\ $$$${which}\:{simplifies}\:{to}\: \\ $$$${q}''\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\rightarrow\:\frac{{d}\left({dq}\right)}{{dx}}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\int\:\frac{{d}\left({dq}\right)}{{dx}}\:=\:\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:\frac{{dq}}{{dx}}\:=\:−\frac{\mathrm{1}}{{x}}\:+{C}_{\mathrm{1}} \: \\ $$$$\int\:{dq}\:=\:\int\:\left(−\frac{\mathrm{1}}{{x}}\:+{C}_{\mathrm{1}} \right)\:{dx} \\ $$$$\rightarrow{q}\:=\:−\mathrm{ln}\:\mid{x}\mid\:+\:{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{y}}{{e}^{\mathrm{3}{x}} }\:=\:−\mathrm{ln}\:\mid{x}\mid\:+\:{C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \\ $$$$\Rightarrow{y}\:=\:−{e}^{\mathrm{3}{x}} .\mathrm{ln}\:\mid{x}\mid\:+\:\left({C}_{\mathrm{1}} {x}+{C}_{\mathrm{2}} \right).{e}^{\mathrm{3}{x}} \\ $$
Answered by Bird last updated on 28/Sep/20
y′′−6y′ +9y =(e^(3x) /x^2 ) h→r^2 −6r +9=0 ⇒(r−3)^2 =0 ⇒ r=3 ⇒y_h =(ax+b)e^(3x) =axe^(3x) +be^(3x) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((xe^(3x) e^(3x) )),(((1+3x)e^(3x) 3e^(3x) ))) =3x e^(6x) −(1+3x)e^(6x) =−e^(6x) ≠0 W_1 = determinant (((o e^(3x) )),(((e^(3x) /x^2 ) 3e^(3x) )))=−(e^(6x) /x^2 ) W_2 = determinant (((xe^(3x) 0)),(((1+3x)e^(3x) (e^(3x) /x^2 ))))=(e^(6x) /x) v_1 =∫ (w_1 /w)dx =∫ −(e^(6x) /(x^2 (−e^(6x) ))) =∫ (dx/x^2 ) =−(1/x) v_2 =∫ (w_2 /w)dx =∫ (e^(6x) /(−xe^(6x) ))dx =−∫ (dx/x) =−ln∣x∣ ⇒ y_p =u_1 v_1 +u_2 v_2 =xe^(3x) (−(1/x)) −e^(3x) ln∣x∣ =−e^(3x) −e^(3x) ln∣x∣ ⇒the general solution is y =axe^(3x) +be^(3x) −(1+ln∣x∣)e^(3x) =(ax+b −1−ln∣x∣)e^(3x)
$${y}''−\mathrm{6}{y}'\:+\mathrm{9}{y}\:=\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} } \\ $$$${h}\rightarrow{r}^{\mathrm{2}} −\mathrm{6}{r}\:+\mathrm{9}=\mathrm{0}\:\Rightarrow\left({r}−\mathrm{3}\right)^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow \\ $$$${r}=\mathrm{3}\:\Rightarrow{y}_{{h}} =\left({ax}+{b}\right){e}^{\mathrm{3}{x}} \:={axe}^{\mathrm{3}{x}} \:+{be}^{\mathrm{3}{x}} \\ $$$$={au}_{\mathrm{1}} \:+{bu}_{\mathrm{2}} \\ $$$${W}\left({u}_{\mathrm{1}} ,{u}_{\mathrm{2}} \right)=\begin{vmatrix}{{xe}^{\mathrm{3}{x}} \:\:\:\:\:\:\:{e}^{\mathrm{3}{x}} }\\{\left(\mathrm{1}+\mathrm{3}{x}\right){e}^{\mathrm{3}{x}} \:\:\:\mathrm{3}{e}^{\mathrm{3}{x}} }\end{vmatrix} \\ $$$$=\mathrm{3}{x}\:{e}^{\mathrm{6}{x}} −\left(\mathrm{1}+\mathrm{3}{x}\right){e}^{\mathrm{6}{x}} \:=−{e}^{\mathrm{6}{x}} \:\neq\mathrm{0} \\ $$$${W}_{\mathrm{1}} =\begin{vmatrix}{{o}\:\:\:\:\:\:\:\:{e}^{\mathrm{3}{x}} }\\{\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} }\:\:\:\:\:\:\mathrm{3}{e}^{\mathrm{3}{x}} }\end{vmatrix}=−\frac{{e}^{\mathrm{6}{x}} }{{x}^{\mathrm{2}} } \\ $$$${W}_{\mathrm{2}} =\begin{vmatrix}{{xe}^{\mathrm{3}{x}} \:\:\:\:\:\:\:\:\mathrm{0}}\\{\left(\mathrm{1}+\mathrm{3}{x}\right){e}^{\mathrm{3}{x}} \:\:\:\frac{{e}^{\mathrm{3}{x}} }{{x}^{\mathrm{2}} }}\end{vmatrix}=\frac{{e}^{\mathrm{6}{x}} }{{x}} \\ $$$${v}_{\mathrm{1}} =\int\:\frac{{w}_{\mathrm{1}} }{{w}}{dx}\:=\int\:−\frac{{e}^{\mathrm{6}{x}} }{{x}^{\mathrm{2}} \left(−{e}^{\mathrm{6}{x}} \right)} \\ $$$$=\int\:\frac{{dx}}{{x}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{{x}} \\ $$$${v}_{\mathrm{2}} =\int\:\frac{{w}_{\mathrm{2}} }{{w}}{dx}\:=\int\:\frac{{e}^{\mathrm{6}{x}} }{−{xe}^{\mathrm{6}{x}} }{dx} \\ $$$$=−\int\:\frac{{dx}}{{x}}\:=−{ln}\mid{x}\mid\:\Rightarrow \\ $$$${y}_{{p}} ={u}_{\mathrm{1}} {v}_{\mathrm{1}} \:+{u}_{\mathrm{2}} {v}_{\mathrm{2}} \\ $$$$={xe}^{\mathrm{3}{x}} \left(−\frac{\mathrm{1}}{{x}}\right)\:−{e}^{\mathrm{3}{x}} {ln}\mid{x}\mid \\ $$$$=−{e}^{\mathrm{3}{x}} −{e}^{\mathrm{3}{x}} {ln}\mid{x}\mid\:\Rightarrow{the}\:{general} \\ $$$${solution}\:{is} \\ $$$${y}\:={axe}^{\mathrm{3}{x}} \:+{be}^{\mathrm{3}{x}} −\left(\mathrm{1}+{ln}\mid{x}\mid\right){e}^{\mathrm{3}{x}} \\ $$$$=\left({ax}+{b}\:−\mathrm{1}−{ln}\mid{x}\mid\right){e}^{\mathrm{3}{x}} \\ $$$$ \\ $$

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