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Question Number 119233 by bemath last updated on 23/Oct/20

(D^2 −5D+6)y = e^(3x)

$$\:\left({D}^{\mathrm{2}} −\mathrm{5}{D}+\mathrm{6}\right){y}\:=\:{e}^{\mathrm{3}{x}} \\ $$

Answered by benjo_mathlover last updated on 23/Oct/20

(D−2I)(D−3I)(y) = e^(3x) ; I(y)=y let (D−3I)(y) = z ⇒ (D−2I)(z) = e^(3x) ⇒ z′ −2z = e^(3x) ; (e^(−2x) )(z′−2z)= e^x ⇒e^(−2x) .z′−2e^(−2x) .z = e^x ⇒(e^(−2x) .z)′ = e^x ⇒ e^(−2x) .z = ∫ e^x dx ⇒e^(−2x) .z = e^x + A; z = e^(3x) +Ae^(2x) ⇒(D−3I) y = e^(3x) +Ae^(2x) ⇒ y′−3y = e^(3x) +Ae^(2x) ⇒(e^(−3x) )(y′−3y) = 1+Ae^(−x) ⇒ (e^(−3x) .y)′ = 1+Ae^(−x) ⇒ e^(−3x) .y = x−Ae^(−x) +B ⇒ y = xe^(3x) −Ae^(2x) +Be^(3x) .

$$\:\left({D}−\mathrm{2}{I}\right)\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{e}^{\mathrm{3}{x}} \:;\:{I}\left({y}\right)={y} \\ $$$${let}\:\left({D}−\mathrm{3}{I}\right)\left({y}\right)\:=\:{z} \\ $$$$\Rightarrow\:\left({D}−\mathrm{2}{I}\right)\left({z}\right)\:=\:{e}^{\mathrm{3}{x}} \\ $$$$\Rightarrow\:{z}'\:−\mathrm{2}{z}\:=\:{e}^{\mathrm{3}{x}} \:;\:\left({e}^{−\mathrm{2}{x}} \right)\left({z}'−\mathrm{2}{z}\right)=\:{e}^{{x}} \\ $$$$\Rightarrow{e}^{−\mathrm{2}{x}} .{z}'−\mathrm{2}{e}^{−\mathrm{2}{x}} .{z}\:=\:{e}^{{x}} \\ $$$$\Rightarrow\left({e}^{−\mathrm{2}{x}} .{z}\right)'\:=\:{e}^{{x}} \\ $$$$\Rightarrow\:{e}^{−\mathrm{2}{x}} .{z}\:=\:\int\:{e}^{{x}} \:{dx}\: \\ $$$$\Rightarrow{e}^{−\mathrm{2}{x}} .{z}\:=\:{e}^{{x}} \:+\:{A};\:{z}\:=\:{e}^{\mathrm{3}{x}} +{Ae}^{\mathrm{2}{x}} \\ $$$$\Rightarrow\left({D}−\mathrm{3}{I}\right)\:{y}\:=\:{e}^{\mathrm{3}{x}} +{Ae}^{\mathrm{2}{x}} \\ $$$$\Rightarrow\:{y}'−\mathrm{3}{y}\:=\:{e}^{\mathrm{3}{x}} +{Ae}^{\mathrm{2}{x}} \\ $$$$\Rightarrow\left({e}^{−\mathrm{3}{x}} \right)\left({y}'−\mathrm{3}{y}\right)\:=\:\mathrm{1}+{Ae}^{−{x}} \\ $$$$\Rightarrow\:\left({e}^{−\mathrm{3}{x}} .{y}\right)'\:=\:\mathrm{1}+{Ae}^{−{x}} \\ $$$$\Rightarrow\:{e}^{−\mathrm{3}{x}} .{y}\:=\:{x}−{Ae}^{−{x}} +{B} \\ $$$$\Rightarrow\:{y}\:=\:{xe}^{\mathrm{3}{x}} −{Ae}^{\mathrm{2}{x}} +{Be}^{\mathrm{3}{x}} \:. \\ $$$$ \\ $$

Commented by bemath last updated on 23/Oct/20

great

$${great} \\ $$

Answered by TANMAY PANACEA last updated on 23/Oct/20

C.F y=e^(mx) m^2 −5m+6=0 (m−2)(m−3)=0→m=2,3 C.F=Ae^(2x) +Be^(3x) P.I y=(e^(3x) /((D−3)(D−2))) y=(e^(3x) /((3−2)))×(1/((D+3−3)))=(e^(3x) /1)×x =(e^(3x) /1)×x complete solution y=Ae^(2x) +Be^(3x) +((xe^(3x) )/1)

$${C}.{F} \\ $$$${y}={e}^{{mx}} \\ $$$${m}^{\mathrm{2}} −\mathrm{5}{m}+\mathrm{6}=\mathrm{0} \\ $$$$\left({m}−\mathrm{2}\right)\left({m}−\mathrm{3}\right)=\mathrm{0}\rightarrow{m}=\mathrm{2},\mathrm{3} \\ $$$${C}.{F}={Ae}^{\mathrm{2}{x}} +{Be}^{\mathrm{3}{x}} \\ $$$${P}.{I} \\ $$$${y}=\frac{{e}^{\mathrm{3}{x}} }{\left({D}−\mathrm{3}\right)\left({D}−\mathrm{2}\right)} \\ $$$$\boldsymbol{{y}}=\frac{{e}^{\mathrm{3}{x}} }{\left(\mathrm{3}−\mathrm{2}\right)}×\frac{\mathrm{1}}{\left({D}+\mathrm{3}−\mathrm{3}\right)}=\frac{{e}^{\mathrm{3}{x}} }{\mathrm{1}}×{x} \\ $$$$=\frac{{e}^{\mathrm{3}{x}} }{\mathrm{1}}×{x} \\ $$$${complete}\:{solution} \\ $$$${y}={Ae}^{\mathrm{2}{x}} +{Be}^{\mathrm{3}{x}} +\frac{{xe}^{\mathrm{3}{x}} }{\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 24/Oct/20

y^(′′) −5y^′ +6y =e^(3x) h→r^2 −5r+6 =0 →Δ=25−24=1 ⇒r_1 =((5+1)/2)=3 r_2 =((5−3)/2)=1 ⇒y_h =ae^x +be^(3x) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((e^x e^(3x) )),((e^x 3e^(3x) )))=3e^(4x) −e^(4x) =2e^(4x) ≠0 W_1 = determinant (((0 e^(3x) )),((e^(3x) 3e^(3x) )))=−e^(6x) W_2 = determinant (((e^x 0)),((e^x e^(3x) )))=e^(4x) v_1 =∫ (W_1 /W)dx =∫ ((−e^(6x) )/(2e^(4x) ))dx =−(1/2)∫e^(2x) dx =−(1/4)e^(2x) v_2 =∫ (W_2 /W)dx =∫ (e^(4x) /(2e^(4x) ))dx =(x/2) ⇒ y_p =u_1 v_1 +u_2 v_2 =e^x (−(1/4)e^(2x) ) +e^(3x) ((x/2)) =−(1/4)e^(3x) +(x/2)e^(3x) ⇒ the general solution is y =y_p +y_h ⇒y =((x/2)−(1/4))e^(3x) +ae^x +b e^(3x)

$$\mathrm{y}^{''} −\mathrm{5y}^{'} +\mathrm{6y}\:=\mathrm{e}^{\mathrm{3x}} \\ $$$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{5r}+\mathrm{6}\:=\mathrm{0}\:\rightarrow\Delta=\mathrm{25}−\mathrm{24}=\mathrm{1}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =\frac{\mathrm{5}+\mathrm{1}}{\mathrm{2}}=\mathrm{3} \\ $$$$\mathrm{r}_{\mathrm{2}} =\frac{\mathrm{5}−\mathrm{3}}{\mathrm{2}}=\mathrm{1}\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\mathrm{x}} \:+\mathrm{be}^{\mathrm{3x}} \:=\mathrm{au}_{\mathrm{1}} +\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\mathrm{e}^{\mathrm{3x}} }\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\mathrm{3e}^{\mathrm{3x}} }\end{vmatrix}=\mathrm{3e}^{\mathrm{4x}} −\mathrm{e}^{\mathrm{4x}} =\mathrm{2e}^{\mathrm{4x}} \:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{3x}} }\\{\mathrm{e}^{\mathrm{3x}} \:\:\:\:\:\:\mathrm{3e}^{\mathrm{3x}} }\end{vmatrix}=−\mathrm{e}^{\mathrm{6x}} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{e}^{\mathrm{x}} \:\:\:\:\:\:\:\:\:\mathrm{e}^{\mathrm{3x}} }\end{vmatrix}=\mathrm{e}^{\mathrm{4x}} \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{W}_{\mathrm{1}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\:\frac{−\mathrm{e}^{\mathrm{6x}} }{\mathrm{2e}^{\mathrm{4x}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{e}^{\mathrm{2x}} \mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{\mathrm{2x}} \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{W}_{\mathrm{2}} }{\mathrm{W}}\mathrm{dx}\:=\int\:\frac{\mathrm{e}^{\mathrm{4x}} }{\mathrm{2e}^{\mathrm{4x}} }\mathrm{dx}\:=\frac{\mathrm{x}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} =\mathrm{e}^{\mathrm{x}} \left(−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{\mathrm{2x}} \right)\:+\mathrm{e}^{\mathrm{3x}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{e}^{\mathrm{3x}} \:+\frac{\mathrm{x}}{\mathrm{2}}\mathrm{e}^{\mathrm{3x}} \:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{y}\:=\mathrm{y}_{\mathrm{p}} \:+\mathrm{y}_{\mathrm{h}} \\ $$$$\Rightarrow\mathrm{y}\:=\left(\frac{\mathrm{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{e}^{\mathrm{3x}} \:+\mathrm{ae}^{\mathrm{x}} \:+\mathrm{b}\:\mathrm{e}^{\mathrm{3x}} \\ $$

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