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Question Number 187046 by Humble last updated on 12/Feb/23

find the general solution for the following system of equations ((dx_1 /dt))=2x_1 +2x_2 ((dx_2 /dt))=x_1 +3x_2

findthegeneralsolutionforthefollowingsystemofequations(dx1dt)=2x1+2x2(dx2dt)=x1+3x2

Answered by mr W last updated on 13/Feb/23

[((2−λ),2),(1,(3−λ)) ]=0 (2−λ)(3−λ)−1×2=0 (λ−1)(λ−4)=0 ⇒λ=1, 4 [((2−1),2),(1,(3−1)) ] [(v_1 ),(v_2 ) ]=0 ⇒ [((−2)),(1) ] [((2−4),2),(1,(3−4)) ] [(v_1 ),(v_2 ) ]=0 ⇒ [(1),(1) ] [(x_1 ),(x_2 ) ]=C_1 e^t [((−2)),(1) ]+C_2 e^(4t) [(1),(1) ] or x_1 =−2C_1 e^t +C_2 e^(4t) x_2 =C_1 e^t +C_2 e^(4t)

[2−λ213−λ]=0(2−λ)(3−λ)−1×2=0(λ−1)(λ−4)=0⇒λ=1,4[2−1213−1][v1v2]=0⇒[−21][2−4213−4][v1v2]=0⇒[11][x1x2]=C1et[−21]+C2e4t[11]orx1=−2C1et+C2e4tx2=C1et+C2e4t

Commented by Humble last updated on 13/Feb/23

Much thanks sir

Muchthankssir

Commented by Humble last updated on 13/Feb/23

Commented by mr W last updated on 13/Feb/23

are you sure? is 1×(−2)+2×1=0 correct ? (what i did) or is 1×(−(1/2))+2×1=0 correct? (what you corrected)

areyousure?is1×(−2)+2×1=0correct?(whatidid)oris1×(−12)+2×1=0correct?(whatyoucorrected)

Commented by Humble last updated on 13/Feb/23

sir, the eigen vector v_1 i think the error is determinant (((2−1 2)),((1 3−1))) => determinant (((1 2)),((1 2))) v_1 =>1c_1 +2c_2 ==> 2c_2 =−1c_(1 ) c_2 =((−1)/2)c_1 for c is arbitraly constant v_1 ==> c_2 =((−1)/2) 1c_1 + 2c_2 ===> 1c_1 = −2c_2 c_1 =−2(((−1)/2))=1 determinant ((1),((−(1/2))))=v_1 or [2 −1]=v_1

sir,theeigenvectorv1ithinktheerroris|2−1213−1|=>|1212|v1=>1c1+2c2==>2c2=−1c1c2=−12c1forcisarbitralyconstantv1==>c2=−121c1+2c2===>1c1=−2c2c1=−2(−12)=1|1−12|=v1or[2−1]=v1

Commented by mr W last updated on 13/Feb/23

my error or your error? we want to find v_1 and v_2 such that

myerrororyourerror?wewanttofindv1andv2suchthat

Commented by mr W last updated on 13/Feb/23

Commented by mr W last updated on 13/Feb/23

i said v_1 =−2, v_1 =1. but you corrected to v_1 =−(1/2), v_2 =1

isaidv1=−2,v1=1.butyoucorrectedtov1=−12,v2=1

Commented by mr W last updated on 13/Feb/23

you can also check my final solution:

youcanalsocheckmyfinalsolution:

Commented by mr W last updated on 13/Feb/23

x_1 =−2C_1 e^t +C_2 e^(4t) x_2 =C_1 e^t +C_2 e^(4t) 2x_1 +2x_2 =−4C_1 e^t +2C_2 e^(4t) +2C_1 e^t +2C_2 e^(4t) =−2C_1 e^t +4C_2 e^(4t) x_1 +3x_2 =−2C_1 e^t +C_2 e^(4t) +3C_1 e^t +3C_2 e^(4t) =C_1 e^t +4C_2 e^(4t) (dx_1 /dt)=−2C_1 e^t +4C_2 e^(4t) =2x_1 +2x_2 ✓ (dx_2 /dt)=C_1 e^t +4C_2 e^(4t) =x_1 +3x_2 ✓

x1=−2C1et+C2e4tx2=C1et+C2e4t2x1+2x2=−4C1et+2C2e4t+2C1et+2C2e4t=−2C1et+4C2e4tx1+3x2=−2C1et+C2e4t+3C1et+3C2e4t=C1et+4C2e4tdx1dt=−2C1et+4C2e4t=2x1+2x2✓dx2dt=C1et+4C2e4t=x1+3x2✓

Commented by Humble last updated on 14/Feb/23

all clear, my apology. sir

allclear,myapology.sir

Commented by mr W last updated on 14/Feb/23

alright sir!

alrightsir!

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