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Question Number 201654 by mokys last updated on 10/Dec/23
Answered by aleks041103 last updated on 10/Dec/23
First part: M= ((4,3),(1,(−2)) ) eigenvals: det(M−xI)= determinant (((4−x),3),(1,(−2−x)))=0 ⇒(x−4)(x+2)−3=0 x^2 −2x−11=0 x_(1,2) =((2±(√(4+4.11)))/2)=1±2(√3) eigenvecs: 1) ker(M−x_1 I)=? (((3−2(√3)),3),(1,(−3−2(√3))) ) ((p),(q) ) = ((0),(0) ) ⇒p−(3+2(√3))q=0⇒ ((p),(q) ) = (((3+2(√3))),(1) ) q ⇒ker(M−x_1 I)=span( (((3+2(√3))),(1) ))=span(v_1 ) 2) ker(M−x_2 I)=? (((3+2(√3)),3),(1,(−3+2(√3))) ) ((p),(q) ) = ((0),(0) ) ⇒p−(3−2(√3))q=0⇒ ((p),(q) ) = (((3−2(√3))),(1) ) q ⇒ker(M−x_2 I)=span( (((3−2(√3))),(1) ))=span(v_2 ) Eigen representation of M M=P D P^( −1) D= (((1+2(√3)),0),(0,(1−2(√3))) ) P =(v_1 v_2 )= (((3+2(√3)),(3−2(√3))),(1,1) ), detP =4(√3) ⇒P^( −1) =(1/(detP)) ((1,(2(√3)−3)),((−1),(2(√3)+3)) ) P^( −1) =(1/(4(√3))) ((1,(2(√3)−3)),((−1),(2(√3)+3)) ) ⇒M=PDP^( −1) = =(1/(4(√3))) (((3+2(√3)),(3−2(√3))),(1,1) ) (((1+2(√3)),0),(0,(1−2(√3))) ) ((1,(2(√3)−3)),((−1),(2(√3)+3)) )
$${First}\:{part}: \\ $$$${M}=\begin{pmatrix}{\mathrm{4}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{2}}\end{pmatrix}\: \\ $$$${eigenvals}: \\ $$$${det}\left({M}−{xI}\right)=\begin{vmatrix}{\mathrm{4}−{x}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{2}−{x}}\end{vmatrix}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)−\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{11}=\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}.\mathrm{11}}}{\mathrm{2}}=\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$${eigenvecs}: \\ $$$$\left.\mathrm{1}\right)\:{ker}\left({M}−{x}_{\mathrm{1}} {I}\right)=? \\ $$$$\begin{pmatrix}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\end{pmatrix}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\Rightarrow{p}−\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\right){q}=\mathrm{0}\Rightarrow\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}\end{pmatrix}\:{q} \\ $$$$\Rightarrow{ker}\left({M}−{x}_{\mathrm{1}} {I}\right)={span}\left(\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}\end{pmatrix}\right)={span}\left({v}_{\mathrm{1}} \right) \\ $$$$\left.\mathrm{2}\right)\:{ker}\left({M}−{x}_{\mathrm{2}} {I}\right)=? \\ $$$$\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{3}}\\{\mathrm{1}}&{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}\end{pmatrix}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\Rightarrow{p}−\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\right){q}=\mathrm{0}\Rightarrow\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}\end{pmatrix}\:{q} \\ $$$$\Rightarrow{ker}\left({M}−{x}_{\mathrm{2}} {I}\right)={span}\left(\begin{pmatrix}{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}\end{pmatrix}\right)={span}\left({v}_{\mathrm{2}} \right) \\ $$$$ \\ $$$${Eigen}\:{representation}\:{of}\:{M} \\ $$$${M}={P}\:{D}\:{P}^{\:−\mathrm{1}} \\ $$$${D}=\begin{pmatrix}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\end{pmatrix} \\ $$$${P}\:=\left({v}_{\mathrm{1}} \:\:{v}_{\mathrm{2}} \right)=\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix},\:{detP}\:=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{P}^{\:−\mathrm{1}} =\frac{\mathrm{1}}{{detP}}\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}\end{pmatrix}\: \\ $$$${P}^{\:−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}\end{pmatrix} \\ $$$$\Rightarrow{M}={PDP}^{\:−\mathrm{1}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\:\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}\end{pmatrix} \\ $$$$ \\ $$$$ \\ $$
Answered by aleks041103 last updated on 10/Dec/23
Second part: (dv^→ /dt)=Mv^→ +f^→ (t) M=PDP^( −1) ⇒(dv^→ /dt)=PDP^( −1) v^→ +f^→ (t) ∣ P^( −1) (∙) (d/dt)(P^( −1) v^→ )=D(P^( −1) v^→ )+(P^( −1) f^→ (t)) let P^( −1) v^→ =u^→ = ((u_1 ),(u_2 ) ) and P^( −1) f^→ (t)=g^→ (t)= (((g_1 (t))),((g_2 (t))) ) and D= ((λ_1 ,0),(0,λ_2 ) ) then (du^→ /dt)=Du^→ +g^→ ⇔ { ((u_1 ′=λ_1 u_1 +g_1 )),((u_2 ′=λ_2 u_2 +g_2 )) :} so this simplifies to solving two simple independent first order ODEs. two Then: ((x),(y) ) = P ((u_1 ),(u_2 ) )
$${Second}\:{part}: \\ $$$$\frac{{d}\overset{\rightarrow} {{v}}}{{dt}}={M}\overset{\rightarrow} {{v}}+\overset{\rightarrow} {{f}}\left({t}\right) \\ $$$${M}={PDP}^{\:−\mathrm{1}} \\ $$$$\Rightarrow\frac{{d}\overset{\rightarrow} {{v}}}{{dt}}={PDP}^{\:−\mathrm{1}} \overset{\rightarrow} {{v}}+\overset{\rightarrow} {{f}}\left({t}\right)\:\:\:\:\:\:\:\:\:\:\mid\:{P}^{\:−\mathrm{1}} \left(\centerdot\right) \\ $$$$\frac{{d}}{{dt}}\left({P}^{\:−\mathrm{1}} \overset{\rightarrow} {{v}}\right)={D}\left({P}^{\:−\mathrm{1}} \overset{\rightarrow} {{v}}\right)+\left({P}^{\:−\mathrm{1}} \overset{\rightarrow} {{f}}\left({t}\right)\right) \\ $$$${let}\:{P}^{\:−\mathrm{1}} \overset{\rightarrow} {{v}}=\overset{\rightarrow} {{u}}=\begin{pmatrix}{{u}_{\mathrm{1}} }\\{{u}_{\mathrm{2}} }\end{pmatrix}\:{and}\:{P}^{\:−\mathrm{1}} \overset{\rightarrow} {{f}}\left({t}\right)=\overset{\rightarrow} {{g}}\left({t}\right)=\begin{pmatrix}{{g}_{\mathrm{1}} \left({t}\right)}\\{{g}_{\mathrm{2}} \left({t}\right)}\end{pmatrix} \\ $$$${and}\:{D}=\begin{pmatrix}{\lambda_{\mathrm{1}} }&{\mathrm{0}}\\{\mathrm{0}}&{\lambda_{\mathrm{2}} }\end{pmatrix} \\ $$$${then} \\ $$$$\frac{{d}\overset{\rightarrow} {{u}}}{{dt}}={D}\overset{\rightarrow} {{u}}+\overset{\rightarrow} {{g}}\:\Leftrightarrow\:\begin{cases}{{u}_{\mathrm{1}} '=\lambda_{\mathrm{1}} {u}_{\mathrm{1}} +{g}_{\mathrm{1}} }\\{{u}_{\mathrm{2}} '=\lambda_{\mathrm{2}} {u}_{\mathrm{2}} +{g}_{\mathrm{2}} }\end{cases} \\ $$$${so}\:{this}\:{simplifies}\:{to}\:{solving}\:{two} \\ $$$${simple}\:{independent}\:{first}\:{order}\:{ODEs}. \\ $$$${two} \\ $$$${Then}: \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:{P}\:\begin{pmatrix}{{u}_{\mathrm{1}} }\\{{u}_{\mathrm{2}} }\end{pmatrix} \\ $$
Answered by aleks041103 last updated on 10/Dec/23
P^( −1) =(1/(4(√3))) ((1,(2(√3)−3)),((−1),(2(√3)+3)) ) g=P^( −1) f=(1/(4(√3))) ((1,(2(√3)−3)),((−1),(2(√3)+3)) ) ((t^2 ),(e^t ) ) = = (((((2(√3)−3)e^t +t^2 )/(4(√3)))),((((2(√(3+))3)e^t −t^2 )/(4(√3)))) ) = ((g_1 ),(g_2 ) ) D= (((1+2(√3)),0),(0,(1−2(√3))) ) ⇒ { ((λ_1 =1+2(√3))),((λ_2 =1−2(√3))) :} ⇒ { ((u_1 ′=(1+2(√3))u_1 +(((2(√3)−3)e^t +t^2 )/(4(√3))))),((u_2 ′=(1−2(√3))u_2 +(((2(√3)+3)e^t −t^2 )/(4(√3))))) :}
$${P}^{\:−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}\end{pmatrix} \\ $$$${g}={P}^{\:−\mathrm{1}} {f}=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\begin{pmatrix}{\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}}\\{−\mathrm{1}}&{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{t}^{\mathrm{2}} }\\{{e}^{{t}} }\end{pmatrix}\:= \\ $$$$=\begin{pmatrix}{\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right){e}^{{t}} +{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}}}\\{\frac{\left(\mathrm{2}\sqrt{\mathrm{3}+}\mathrm{3}\right){e}^{{t}} −{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}}}\end{pmatrix}\:\:=\:\begin{pmatrix}{{g}_{\mathrm{1}} }\\{{g}_{\mathrm{2}} }\end{pmatrix} \\ $$$${D}=\begin{pmatrix}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\end{pmatrix}\:\Rightarrow\begin{cases}{\lambda_{\mathrm{1}} =\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}}\\{\lambda_{\mathrm{2}} =\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{u}_{\mathrm{1}} '=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{1}} +\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right){e}^{{t}} +{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}}}\\{{u}_{\mathrm{2}} '=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{2}} +\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\right){e}^{{t}} −{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}}}\end{cases} \\ $$
Answered by aleks041103 last updated on 10/Dec/23
u_1 ′=(1+2(√3))u_1 +(((2(√3)−3)e^t +t^2 )/(4(√3))) partial soln: u_1 =Ae^t +Bt^2 +Ct+D ⇒Ae^t +2Bt+C=(1+2(√3))(Ae^t +Bt^2 +Ct+D)+((2−(√3))/4)e^t +(1/(4(√3)))t^2 ⇒ { ((A=(1+2(√3))A+((2−(√3))/4))),((0=(1+2(√3))B+(1/(4(√3))))),((2B=(1+2(√3))C)),((C=(1+2(√3))D)) :} ⇒A=(((√3)−2)/(8(√3)));B=−(1/(4(√3)(1+2(√3)))); C=−(1/(2(√3)(1+2(√3))^2 )); D=−(1/(2(√3)(1+2(√3))^3 )) ⇒u_1 =(((√3)−2)/(8(√3)))e^t −(([(1+2(√3))t]^2 +2[(1+2(√3))t]+2)/(4(√3)(1+2(√3))^3 )) homogenuous soln: u_1 ′=(1+2(√3))u_1 ⇒u_1 =C_1 e^((1+2(√3))t) ⇒u_1 =C_1 e^((1+2(√3))t) +(((√3)−2)/(8(√3)))e^t −(([(1+2(√3))t]^2 +2[(1+2(√3))t]+2)/(4(√3)(1+2(√3))^3 ))
$${u}_{\mathrm{1}} '=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{1}} +\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right){e}^{{t}} +{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$${partial}\:{soln}: \\ $$$${u}_{\mathrm{1}} ={Ae}^{{t}} +{Bt}^{\mathrm{2}} +{Ct}+{D} \\ $$$$\Rightarrow{Ae}^{{t}} +\mathrm{2}{Bt}+{C}=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)\left({Ae}^{{t}} +{Bt}^{\mathrm{2}} +{Ct}+{D}\right)+\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}{e}^{{t}} +\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}{t}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{A}=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){A}+\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}}}\\{\mathrm{0}=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){B}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}}\\{\mathrm{2}{B}=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){C}}\\{{C}=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){D}}\end{cases} \\ $$$$\Rightarrow{A}=\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{8}\sqrt{\mathrm{3}}};{B}=−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)};\:{C}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }; \\ $$$${D}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{u}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$${homogenuous}\:{soln}: \\ $$$${u}_{\mathrm{1}} '=\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{1}} \Rightarrow{u}_{\mathrm{1}} ={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} \\ $$$$ \\ $$$$\Rightarrow{u}_{\mathrm{1}} ={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$
Answered by aleks041103 last updated on 10/Dec/23
u_2 ′=(1−2(√3))u_2 +(((2(√3)+3)e^t −t^2 )/(4(√3))) partial soln: u_2 =Ae^t +Bt^2 +Ct+D ⇒Ae^t +2Bt+C=(1−2(√3))(Ae^t +Bt^2 +Ct+D)+((2+(√3))/4)e^t −(1/(4(√3)))t^2 ⇒ { ((A=(1−2(√3))A+((2+(√3))/4))),((0=(1−2(√3))B−(1/(4(√3))))),((2B=(1−2(√3))C)),((C=(1−2(√3))D)) :} ⇒A=((2+(√3))/(8(√3)));B=(1/(4(√3)(1−2(√3)))); C=(1/(2(√3)(1−2(√3))^2 )); D=(1/(2(√3)(1−2(√3))^3 )) ⇒u_2 =((2+(√3))/(8(√3)))e^t +(([(1−2(√3))t]^2 +2[(1−2(√3))t]+2)/(4(√3)(1−2(√3))^3 )) homogenuous soln: u_2 ′=(1−2(√3))u_2 ⇒u_2 =C_2 e^((1−2(√3))t) ⇒u_2 =C_2 e^((1−2(√3))t) +((2+(√3))/(8(√3)))e^t +(([(1−2(√3))t]^2 +2[(1−2(√3))t]+2)/(4(√3)(1−2(√3))^3 ))
$${u}_{\mathrm{2}} '=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{2}} +\frac{\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\right){e}^{{t}} −{t}^{\mathrm{2}} }{\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$${partial}\:{soln}: \\ $$$${u}_{\mathrm{2}} ={Ae}^{{t}} +{Bt}^{\mathrm{2}} +{Ct}+{D} \\ $$$$\Rightarrow{Ae}^{{t}} +\mathrm{2}{Bt}+{C}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)\left({Ae}^{{t}} +{Bt}^{\mathrm{2}} +{Ct}+{D}\right)+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}{e}^{{t}} −\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}{t}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{A}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){A}+\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}}\\{\mathrm{0}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){B}−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}}\\{\mathrm{2}{B}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){C}}\\{{C}=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){D}}\end{cases} \\ $$$$\Rightarrow{A}=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{8}\sqrt{\mathrm{3}}};{B}=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)};\:{C}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }; \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{u}_{\mathrm{2}} =\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} +\frac{\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$$${homogenuous}\:{soln}: \\ $$$${u}_{\mathrm{2}} '=\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){u}_{\mathrm{2}} \Rightarrow{u}_{\mathrm{2}} ={C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} \\ $$$$ \\ $$$$\Rightarrow{u}_{\mathrm{2}} ={C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} +\frac{\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} } \\ $$
Answered by aleks041103 last updated on 10/Dec/23
Finally ((x),(y) ) = P ((u_1 ),(u_2 ) ) P = (((3+2(√3)),(3−2(√3))),(1,1) ) ((u_1 ),(u_2 ) ) = (((C_1 e^((1+2(√3))t) +(((√3)−2)/(8(√3)))e^t −(([(1+2(√3))t]^2 +2[(1+2(√3))t]+2)/(4(√3)(1+2(√3))^3 )))),((C_2 e^((1−2(√3))t) +((2+(√3))/(8(√3)))e^t +(([(1−2(√3))t]^2 +2[(1−2(√3))t]+2)/(4(√3)(1−2(√3))^3 )))) ) ⇒x=3(u_1 +u_2 )+2(√3)(u_1 −u_2 ) y=u_1 +u_2 ⇒u_1 +u_2 =C_1 e^((1+2(√3))t) +C_2 e^((1−2(√3))t) +(e^t /4)+(([(1−2(√3))t]^2 +2[(1−2(√3))t]+2)/(4(√3)(1−2(√3))^3 ))−(([(1+2(√3))t]^2 +2[(1+2(√3))t]+2)/(4(√3)(1+2(√3))^3 ))= =C_1 e^((1+2(√3))t) +C_2 e^((1−2(√3))t) +(e^t /4)−(t^2 /(11))+((4t)/(121))+((30)/(1331)) u_1 −u_2 =C_1 e^((1+2(√3))t) −C_2 e^((1−2(√3))t) −(e^t /(2(√3)))+(t^2 /(11.2(√3)))−((26t)/(121.2(√3)))−((74)/(1331.2(√3))) ⇒ x=(3+2(√3))C_1 e^((1+2(√3))t) +(3−2(√3))C_2 e^((1−2(√3))t) −(e^t /4)−((2t^2 )/(11))−((14t)/(121))+((16)/(1331)) y=C_1 e^((1+2(√3))t) +C_2 e^((1−2(√3))t) +(e^t /4)−(t^2 /(11))+((4t)/(121))+((30)/(1331))
$${Finally} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:{P}\:\begin{pmatrix}{{u}_{\mathrm{1}} }\\{{u}_{\mathrm{2}} }\end{pmatrix} \\ $$$${P}\:=\:\begin{pmatrix}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}}&{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{u}_{\mathrm{1}} }\\{{u}_{\mathrm{2}} }\end{pmatrix}\:=\:\begin{pmatrix}{{C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }}\\{{C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{8}\sqrt{\mathrm{3}}}{e}^{{t}} +\frac{\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }}\end{pmatrix} \\ $$$$\Rightarrow{x}=\mathrm{3}\left({u}_{\mathrm{1}} +{u}_{\mathrm{2}} \right)+\mathrm{2}\sqrt{\mathrm{3}}\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right) \\ $$$${y}={u}_{\mathrm{1}} +{u}_{\mathrm{2}} \\ $$$$\Rightarrow{u}_{\mathrm{1}} +{u}_{\mathrm{2}} ={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +{C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{{e}^{{t}} }{\mathrm{4}}+\frac{\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]^{\mathrm{2}} +\mathrm{2}\left[\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}\right]+\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }= \\ $$$$={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +{C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{{e}^{{t}} }{\mathrm{4}}−\frac{{t}^{\mathrm{2}} }{\mathrm{11}}+\frac{\mathrm{4}{t}}{\mathrm{121}}+\frac{\mathrm{30}}{\mathrm{1331}} \\ $$$${u}_{\mathrm{1}} −{u}_{\mathrm{2}} ={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} −{C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} −\frac{{e}^{{t}} }{\mathrm{2}\sqrt{\mathrm{3}}}+\frac{{t}^{\mathrm{2}} }{\mathrm{11}.\mathrm{2}\sqrt{\mathrm{3}}}−\frac{\mathrm{26}{t}}{\mathrm{121}.\mathrm{2}\sqrt{\mathrm{3}}}−\frac{\mathrm{74}}{\mathrm{1331}.\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$\Rightarrow \\ $$$${x}=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\right){C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{3}}\right){C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} −\frac{{e}^{{t}} }{\mathrm{4}}−\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{11}}−\frac{\mathrm{14}{t}}{\mathrm{121}}+\frac{\mathrm{16}}{\mathrm{1331}} \\ $$$${y}={C}_{\mathrm{1}} {e}^{\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +{C}_{\mathrm{2}} {e}^{\left(\mathrm{1}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}} +\frac{{e}^{{t}} }{\mathrm{4}}−\frac{{t}^{\mathrm{2}} }{\mathrm{11}}+\frac{\mathrm{4}{t}}{\mathrm{121}}+\frac{\mathrm{30}}{\mathrm{1331}} \\ $$

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