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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)) )

Firstpart:M=(431−2)eigenvals:det(M−xI)=|4−x31−2−x|=0⇒(x−4)(x+2)−3=0x2−2x−11=0x1,2=2±4+4.112=1±23eigenvecs:1)ker(M−x1I)=?(3−2331−3−23)(pq)=(00)⇒p−(3+23)q=0⇒(pq)=(3+231)q⇒ker(M−x1I)=span((3+231))=span(v1)2)ker(M−x2I)=?(3+2331−3+23)(pq)=(00)⇒p−(3−23)q=0⇒(pq)=(3−231)q⇒ker(M−x2I)=span((3−231))=span(v2)EigenrepresentationofMM=PDP−1D=(1+23001−23)P=(v1v2)=(3+233−2311),detP=43⇒P−1=1detP(123−3−123+3)P−1=143(123−3−123+3)⇒M=PDP−1==143(3+233−2311)(1+23001−23)(123−3−123+3)

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 ) )

Secondpart:dv→dt=Mv→+f→(t)M=PDP−1⇒dv→dt=PDP−1v→+f→(t)∣P−1(⋅)ddt(P−1v→)=D(P−1v→)+(P−1f→(t))letP−1v→=u→=(u1u2)andP−1f→(t)=g→(t)=(g1(t)g2(t))andD=(λ100λ2)thendu→dt=Du→+g→⇔{u1′=λ1u1+g1u2′=λ2u2+g2sothissimplifiestosolvingtwosimpleindependentfirstorderODEs.twoThen:(xy)=P(u1u2)

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−1=143(123−3−123+3)g=P−1f=143(123−3−123+3)(t2et)==((23−3)et+t243(23+3)et−t243)=(g1g2)D=(1+23001−23)⇒{λ1=1+23λ2=1−23⇒{u1′=(1+23)u1+(23−3)et+t243u2′=(1−23)u2+(23+3)et−t243

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 ))

u1′=(1+23)u1+(23−3)et+t243partialsoln:u1=Aet+Bt2+Ct+D⇒Aet+2Bt+C=(1+23)(Aet+Bt2+Ct+D)+2−34et+143t2⇒{A=(1+23)A+2−340=(1+23)B+1432B=(1+23)CC=(1+23)D⇒A=3−283;B=−143(1+23);C=−123(1+23)2;D=−123(1+23)3⇒u1=3−283et−[(1+23)t]2+2[(1+23)t]+243(1+23)3homogenuoussoln:u1′=(1+23)u1⇒u1=C1e(1+23)t⇒u1=C1e(1+23)t+3−283et−[(1+23)t]2+2[(1+23)t]+243(1+23)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 ))

u2′=(1−23)u2+(23+3)et−t243partialsoln:u2=Aet+Bt2+Ct+D⇒Aet+2Bt+C=(1−23)(Aet+Bt2+Ct+D)+2+34et−143t2⇒{A=(1−23)A+2+340=(1−23)B−1432B=(1−23)CC=(1−23)D⇒A=2+383;B=143(1−23);C=123(1−23)2;D=123(1−23)3⇒u2=2+383et+[(1−23)t]2+2[(1−23)t]+243(1−23)3homogenuoussoln:u2′=(1−23)u2⇒u2=C2e(1−23)t⇒u2=C2e(1−23)t+2+383et+[(1−23)t]2+2[(1−23)t]+243(1−23)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(xy)=P(u1u2)P=(3+233−2311)(u1u2)=(C1e(1+23)t+3−283et−[(1+23)t]2+2[(1+23)t]+243(1+23)3C2e(1−23)t+2+383et+[(1−23)t]2+2[(1−23)t]+243(1−23)3)⇒x=3(u1+u2)+23(u1−u2)y=u1+u2⇒u1+u2=C1e(1+23)t+C2e(1−23)t+et4+[(1−23)t]2+2[(1−23)t]+243(1−23)3−[(1+23)t]2+2[(1+23)t]+243(1+23)3==C1e(1+23)t+C2e(1−23)t+et4−t211+4t121+301331u1−u2=C1e(1+23)t−C2e(1−23)t−et23+t211.23−26t121.23−741331.23⇒x=(3+23)C1e(1+23)t+(3−23)C2e(1−23)t−et4−2t211−14t121+161331y=C1e(1+23)t+C2e(1−23)t+et4−t211+4t121+301331

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