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find-the-general-solution-for-the-following-system-of-equations-dx-1-dt-2x-1-2x-2-dx-2-dt-x-1-3x-2-

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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
$${find}\:\:\:{the}\:{general}\:\:{solution}\:{for}\:\:{the}\:{following} \\ $$$${system}\:{of}\:{equations} \\ $$$$\left(\frac{{dx}_{\mathrm{1}} }{{dt}}\right)=\mathrm{2}{x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} \\ $$$$\left(\frac{{dx}_{\mathrm{2}} }{{dt}}\right)={x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} \\ $$$$ \\ $$
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)
$$\begin{bmatrix}{\mathrm{2}−\lambda}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}−\lambda}\end{bmatrix}=\mathrm{0} \\ $$$$\left(\mathrm{2}−\lambda\right)\left(\mathrm{3}−\lambda\right)−\mathrm{1}×\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda−\mathrm{1}\right)\left(\lambda−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{1},\:\mathrm{4} \\ $$$$\begin{bmatrix}{\mathrm{2}−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}−\mathrm{1}}\end{bmatrix}\begin{bmatrix}{{v}_{\mathrm{1}} }\\{{v}_{\mathrm{2}} }\end{bmatrix}=\mathrm{0} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{2}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}−\mathrm{4}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{3}−\mathrm{4}}\end{bmatrix}\begin{bmatrix}{{v}_{\mathrm{1}} }\\{{v}_{\mathrm{2}} }\end{bmatrix}=\mathrm{0} \\ $$$$\Rightarrow\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{{x}_{\mathrm{1}} }\\{{x}_{\mathrm{2}} }\end{bmatrix}={C}_{\mathrm{1}} {e}^{{t}} \begin{bmatrix}{−\mathrm{2}}\\{\mathrm{1}}\end{bmatrix}+{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix} \\ $$$${or} \\ $$$${x}_{\mathrm{1}} =−\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$${x}_{\mathrm{2}} ={C}_{\mathrm{1}} {e}^{{t}} +{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$
Commented by Humble last updated on 13/Feb/23
Much thanks sir
$${Much}\:\:{thanks}\:\:\:{sir} \\ $$
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)
$${are}\:{you}\:{sure}? \\ $$$${is}\:\mathrm{1}×\left(−\mathrm{2}\right)+\mathrm{2}×\mathrm{1}=\mathrm{0}\:{correct}\:? \\ $$$$\left({what}\:{i}\:{did}\right) \\ $$$${or} \\ $$$${is}\:\mathrm{1}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}×\mathrm{1}=\mathrm{0}\:{correct}? \\ $$$$\left({what}\:{you}\:{corrected}\right) \\ $$
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
$${sir},\:{the}\:{eigen}\:{vector}\:{v}_{\mathrm{1}} \: \\ $$$${i}\:{think}\:{the}\:{error}\:{is} \\ $$$$\begin{vmatrix}{\mathrm{2}−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}−\mathrm{1}}\end{vmatrix}\:=>\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix} \\ $$$$ \\ $$$${v}_{\mathrm{1}} \:=>\mathrm{1}{c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{2}} \:==>\:\:\mathrm{2}{c}_{\mathrm{2}} \:=−\mathrm{1}{c}_{\mathrm{1}\:\:} \:\:{c}_{\mathrm{2}} =\frac{−\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{1}} \\ $$$${for}\:{c}\:{is}\:{arbitraly}\:{constant} \\ $$$${v}_{\mathrm{1}} \:==>\:{c}_{\mathrm{2}} =\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}{c}_{\mathrm{1}} \:+\:\mathrm{2}{c}_{\mathrm{2}} \:===>\:\mathrm{1}{c}_{\mathrm{1}} \:=\:−\mathrm{2}{c}_{\mathrm{2}} \:\:{c}_{\mathrm{1}} =−\mathrm{2}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$$\begin{vmatrix}{\mathrm{1}}\\{−\frac{\mathrm{1}}{\mathrm{2}}}\end{vmatrix}={v}_{\mathrm{1}} \:{or}\:\left[\mathrm{2}\:\:\:\:\:\:−\mathrm{1}\right]={v}_{\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$
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
$${my}\:{error}\:{or}\:{your}\:{error}? \\ $$$${we}\:{want}\:{to}\:{find}\:{v}_{\mathrm{1}} \:{and}\:{v}_{\mathrm{2}} \:{such}\:{that} \\ $$
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
$${i}\:{said}\:{v}_{\mathrm{1}} =−\mathrm{2},\:{v}_{\mathrm{1}} =\mathrm{1}. \\ $$$${but}\:{you}\:{corrected}\:{to} \\ $$$${v}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}},\:{v}_{\mathrm{2}} =\mathrm{1} \\ $$
Commented by mr W last updated on 13/Feb/23
you can also check my final solution:
$${you}\:{can}\:{also}\:{check}\:{my}\:{final}\:{solution}: \\ $$
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 ✓
$${x}_{\mathrm{1}} =−\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$${x}_{\mathrm{2}} ={C}_{\mathrm{1}} {e}^{{t}} +{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$$\mathrm{2}{x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} =−\mathrm{4}{C}_{\mathrm{1}} {e}^{{t}} +\mathrm{2}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} +\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +\mathrm{2}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +\mathrm{4}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$${x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} =−\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} +\mathrm{3}{C}_{\mathrm{1}} {e}^{{t}} +\mathrm{3}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={C}_{\mathrm{1}} {e}^{{t}} +\mathrm{4}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} \\ $$$$\frac{{dx}_{\mathrm{1}} }{{dt}}=−\mathrm{2}{C}_{\mathrm{1}} {e}^{{t}} +\mathrm{4}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} =\mathrm{2}{x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} \:\checkmark \\ $$$$\frac{{dx}_{\mathrm{2}} }{{dt}}={C}_{\mathrm{1}} {e}^{{t}} +\mathrm{4}{C}_{\mathrm{2}} {e}^{\mathrm{4}{t}} ={x}_{\mathrm{1}} +\mathrm{3}{x}_{\mathrm{2}} \:\checkmark \\ $$
Commented by Humble last updated on 14/Feb/23
all clear, my apology. sir
$${all}\:{clear},\:{my}\:{apology}.\:{sir} \\ $$
Commented by mr W last updated on 14/Feb/23
alright sir!
$${alright}\:{sir}! \\ $$

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