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Question Number 207096 by Wuji last updated on 06/May/24
for the given system of ODEs, calculate the eigenvalues and corresponding eigenvectors of the coefficient matrix (dx/dt)=2x+y (dy/dt)=x+2y
$${for}\:{the}\:{given}\:{system}\:{of}\:{ODEs},\:{calculate}\:{the} \\ $$$${eigenvalues}\:{and}\:{corresponding}\:{eigenvectors}\:{of}\:{the}\: \\ $$$${coefficient}\:{matrix} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{2}{x}+{y}\:\:\:\frac{{dy}}{{dt}}={x}+\mathrm{2}{y} \\ $$
Commented by Wuji last updated on 07/May/24
need a helping hand, please
$${need}\:{a}\:{helping}\:{hand},\:{please} \\ $$
Commented by aleks041103 last updated on 07/May/24
the idea is if q= ((x),(y) ) and also then (dq/dt)= (((dx/dt)),((dy/dt)) ) then you can write the linear ODE as (dq/dt)=M^� q where M^� is the coefficient matrix ⇒ in this case the coeff matrix is ((2,1),(1,2) )... try from here yourself
$${the}\:{idea}\:{is}\: \\ $$$${if}\:\:\:\boldsymbol{{q}}=\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:{and}\:{also}\:{then}\:\:\frac{{d}\boldsymbol{{q}}}{{dt}}=\begin{pmatrix}{{dx}/{dt}}\\{{dy}/{dt}}\end{pmatrix} \\ $$$${then}\:{you}\:{can}\:{write}\:{the}\:{linear}\:{ODE}\:{as} \\ $$$$\frac{{d}\boldsymbol{{q}}}{{dt}}=\hat {{M}}\:\boldsymbol{{q}}\: \\ $$$${where}\:\hat {{M}}\:{is}\:{the}\:{coefficient}\:{matrix} \\ $$$$ \\ $$$$\Rightarrow\:{in}\:{this}\:{case}\:{the}\:{coeff}\:{matrix}\:{is} \\ $$$$\begin{pmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}\end{pmatrix}… \\ $$$${try}\:{from}\:{here}\:{yourself} \\ $$
Commented by Wuji last updated on 07/May/24
yes, sir. thank you so much
$$\mathrm{yes},\:\mathrm{sir}.\:\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by mr W last updated on 07/May/24
alternative way: (i)+(ii): ((d(x+y))/dt)=3(x+y) ((d(x+y))/(x+y))=3dt ⇒ln (x+y)=3t+C ⇒x+y=2C_1 e^(3t) ...(I) (i)−(ii): ((d(x−y))/dt)=x−y ((d(x−y))/(x−y))=dt ⇒ln (x−y)=t+C ⇒x−y=2C_2 e^t ...(II) ⇒x=C_1 e^(3t) +C_2 e^t ⇒y=C_1 e^(3t) −C_2 e^t
$${alternative}\:{way}: \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\frac{{d}\left({x}+{y}\right)}{{dt}}=\mathrm{3}\left({x}+{y}\right) \\ $$$$\frac{{d}\left({x}+{y}\right)}{{x}+{y}}=\mathrm{3}{dt} \\ $$$$\Rightarrow\mathrm{ln}\:\left({x}+{y}\right)=\mathrm{3}{t}+{C} \\ $$$$\Rightarrow{x}+{y}=\mathrm{2}{C}_{\mathrm{1}} {e}^{\mathrm{3}{t}} \:\:\:…\left({I}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\frac{{d}\left({x}−{y}\right)}{{dt}}={x}−{y} \\ $$$$\frac{{d}\left({x}−{y}\right)}{{x}−{y}}={dt} \\ $$$$\Rightarrow\mathrm{ln}\:\left({x}−{y}\right)={t}+{C} \\ $$$$\Rightarrow{x}−{y}=\mathrm{2}{C}_{\mathrm{2}} {e}^{{t}} \:\:\:…\left({II}\right) \\ $$$$\Rightarrow{x}={C}_{\mathrm{1}} {e}^{\mathrm{3}{t}} +{C}_{\mathrm{2}} {e}^{{t}} \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} {e}^{\mathrm{3}{t}} −{C}_{\mathrm{2}} {e}^{{t}} \\ $$
Commented by Wuji last updated on 07/May/24
God bless you, sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you},\:\mathrm{sir} \\ $$

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