System of differential equation stability, with initial condition












1












$begingroup$


Consider the following system of differential equations:



$$ dot x(t) = 3x(t) - 2y(t) + 3$$



$$dot y(t) = 2x(t) - 2y(t) - 1$$



(a) Find the steady state of the system and determine its stability.



(b) Find the particular solution with initial conditions $$x(0) = 0.5$$$$
y(0) = 0$$



I find the general solution$$ x(t) = 2C_1 e^{2t}+C_2 e^{-t}-4 $$
$$y(t) = C_1 e^{2t}+2C_2 e^{-t} -4.5$$
I can see that $C_1 = 0$ otherwise the terms explode, so:
$$ x(t) = C_2 e^{-t}-4 $$
$$y(t) = 2C_2 e^{-t} -4.5$$

But now, the initial conditions lead to different particular solutions, I have done the problem several times, and still cannot seem to find out why. Where have I misstepped?






ordinary-differential-equations systems-of-equations economics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
    $endgroup$
    – DaveNine
    Dec 15 '18 at 23:44










  • $begingroup$
    Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
    $endgroup$
    – Justyen77
    Dec 15 '18 at 23:59


















1












$begingroup$


Consider the following system of differential equations:



$$ dot x(t) = 3x(t) - 2y(t) + 3$$



$$dot y(t) = 2x(t) - 2y(t) - 1$$



(a) Find the steady state of the system and determine its stability.



(b) Find the particular solution with initial conditions $$x(0) = 0.5$$$$
y(0) = 0$$



I find the general solution$$ x(t) = 2C_1 e^{2t}+C_2 e^{-t}-4 $$
$$y(t) = C_1 e^{2t}+2C_2 e^{-t} -4.5$$
I can see that $C_1 = 0$ otherwise the terms explode, so:
$$ x(t) = C_2 e^{-t}-4 $$
$$y(t) = 2C_2 e^{-t} -4.5$$

But now, the initial conditions lead to different particular solutions, I have done the problem several times, and still cannot seem to find out why. Where have I misstepped?






ordinary-differential-equations systems-of-equations economics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
    $endgroup$
    – DaveNine
    Dec 15 '18 at 23:44










  • $begingroup$
    Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
    $endgroup$
    – Justyen77
    Dec 15 '18 at 23:59
















1












1








1





$begingroup$


Consider the following system of differential equations:



$$ dot x(t) = 3x(t) - 2y(t) + 3$$



$$dot y(t) = 2x(t) - 2y(t) - 1$$



(a) Find the steady state of the system and determine its stability.



(b) Find the particular solution with initial conditions $$x(0) = 0.5$$$$
y(0) = 0$$



I find the general solution$$ x(t) = 2C_1 e^{2t}+C_2 e^{-t}-4 $$
$$y(t) = C_1 e^{2t}+2C_2 e^{-t} -4.5$$
I can see that $C_1 = 0$ otherwise the terms explode, so:
$$ x(t) = C_2 e^{-t}-4 $$
$$y(t) = 2C_2 e^{-t} -4.5$$

But now, the initial conditions lead to different particular solutions, I have done the problem several times, and still cannot seem to find out why. Where have I misstepped?






ordinary-differential-equations systems-of-equations economics







share|cite|improve this question











$endgroup$




Consider the following system of differential equations:



$$ dot x(t) = 3x(t) - 2y(t) + 3$$



$$dot y(t) = 2x(t) - 2y(t) - 1$$



(a) Find the steady state of the system and determine its stability.



(b) Find the particular solution with initial conditions $$x(0) = 0.5$$$$
y(0) = 0$$



I find the general solution$$ x(t) = 2C_1 e^{2t}+C_2 e^{-t}-4 $$
$$y(t) = C_1 e^{2t}+2C_2 e^{-t} -4.5$$
I can see that $C_1 = 0$ otherwise the terms explode, so:
$$ x(t) = C_2 e^{-t}-4 $$
$$y(t) = 2C_2 e^{-t} -4.5$$

But now, the initial conditions lead to different particular solutions, I have done the problem several times, and still cannot seem to find out why. Where have I misstepped?






ordinary-differential-equations systems-of-equations economics




ordinary-differential-equations systems-of-equations economics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 0:27









user376343

3,3582826




3,3582826










asked Dec 15 '18 at 23:36









Justyen77Justyen77

83




83








  • 2




    $begingroup$
    Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
    $endgroup$
    – DaveNine
    Dec 15 '18 at 23:44










  • $begingroup$
    Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
    $endgroup$
    – Justyen77
    Dec 15 '18 at 23:59
















  • 2




    $begingroup$
    Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
    $endgroup$
    – DaveNine
    Dec 15 '18 at 23:44










  • $begingroup$
    Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
    $endgroup$
    – Justyen77
    Dec 15 '18 at 23:59










2




2




$begingroup$
Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
$endgroup$
– DaveNine
Dec 15 '18 at 23:44




$begingroup$
Your issue is that you should be using those initial conditions to determine your solution, not looking for solutions that are bounded (aka they don't explode).
$endgroup$
– DaveNine
Dec 15 '18 at 23:44












$begingroup$
Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
$endgroup$
– Justyen77
Dec 15 '18 at 23:59






$begingroup$
Ah, I think I understand, is it right if I say: I should use them with the general solution rather than the bounded solutions where $C_1$ is equal to $0$? And thank you!
$endgroup$
– Justyen77
Dec 15 '18 at 23:59












2 Answers
2






active

oldest

votes


















1












$begingroup$

The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 text{ and } 2x-2y-1=0 $$



which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $xi = x+4$ and $eta = y+9/2$. This gives us the system $$ dotxi = 3xi-2eta \ doteta =2xi - 2eta$$



Stability of this system is determined by the matrix $A = pmatrix{3&-2\2&-2}$, which has $text{tr}A=1 > 0$ and $text{det}A= -2 < 0$. This means that the equilibrium at $(xi,eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.



Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.



We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \ C_2 = frac{1}{2}y_0 + frac{9}{4}$$



which means that only initial conditions on the line $y = 2x+frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.



We can clearly see that $0neq 2(0.5)+frac{7}{2}$, so the given initial conditions will not lead to a stable solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:31












  • $begingroup$
    Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
    $endgroup$
    – AlexanderJ93
    Dec 16 '18 at 0:34



















0












$begingroup$

Hint :



It is $x(0) = 1/2$, thus :



$$x(0) = 0 Rightarrow 2C_1 + C_2 -4 = 1/2$$



Also, $y(0) = 0$, thus :



$$y(0) = 0 Rightarrow C_1 + 2C_2- 9/2 = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:24











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2 Answers
2






active

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2 Answers
2






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 text{ and } 2x-2y-1=0 $$



which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $xi = x+4$ and $eta = y+9/2$. This gives us the system $$ dotxi = 3xi-2eta \ doteta =2xi - 2eta$$



Stability of this system is determined by the matrix $A = pmatrix{3&-2\2&-2}$, which has $text{tr}A=1 > 0$ and $text{det}A= -2 < 0$. This means that the equilibrium at $(xi,eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.



Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.



We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \ C_2 = frac{1}{2}y_0 + frac{9}{4}$$



which means that only initial conditions on the line $y = 2x+frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.



We can clearly see that $0neq 2(0.5)+frac{7}{2}$, so the given initial conditions will not lead to a stable solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:31












  • $begingroup$
    Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
    $endgroup$
    – AlexanderJ93
    Dec 16 '18 at 0:34
















1












$begingroup$

The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 text{ and } 2x-2y-1=0 $$



which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $xi = x+4$ and $eta = y+9/2$. This gives us the system $$ dotxi = 3xi-2eta \ doteta =2xi - 2eta$$



Stability of this system is determined by the matrix $A = pmatrix{3&-2\2&-2}$, which has $text{tr}A=1 > 0$ and $text{det}A= -2 < 0$. This means that the equilibrium at $(xi,eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.



Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.



We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \ C_2 = frac{1}{2}y_0 + frac{9}{4}$$



which means that only initial conditions on the line $y = 2x+frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.



We can clearly see that $0neq 2(0.5)+frac{7}{2}$, so the given initial conditions will not lead to a stable solution.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:31












  • $begingroup$
    Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
    $endgroup$
    – AlexanderJ93
    Dec 16 '18 at 0:34














1












1








1





$begingroup$

The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 text{ and } 2x-2y-1=0 $$



which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $xi = x+4$ and $eta = y+9/2$. This gives us the system $$ dotxi = 3xi-2eta \ doteta =2xi - 2eta$$



Stability of this system is determined by the matrix $A = pmatrix{3&-2\2&-2}$, which has $text{tr}A=1 > 0$ and $text{det}A= -2 < 0$. This means that the equilibrium at $(xi,eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.



Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.



We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \ C_2 = frac{1}{2}y_0 + frac{9}{4}$$



which means that only initial conditions on the line $y = 2x+frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.



We can clearly see that $0neq 2(0.5)+frac{7}{2}$, so the given initial conditions will not lead to a stable solution.






share|cite|improve this answer











$endgroup$



The steady state of the system is the point where both derivatives are zero simultaneously, i.e. the intersection of the lines $$ 3x-2y+3=0 text{ and } 2x-2y-1=0 $$



which occurs at the point $(x,y) = (-4,-9/2)$. We make a change of variables so that the equilibrium is at the origin, which should make a homogeneous linear system. Let $xi = x+4$ and $eta = y+9/2$. This gives us the system $$ dotxi = 3xi-2eta \ doteta =2xi - 2eta$$



Stability of this system is determined by the matrix $A = pmatrix{3&-2\2&-2}$, which has $text{tr}A=1 > 0$ and $text{det}A= -2 < 0$. This means that the equilibrium at $(xi,eta) = (0,0)$ is a saddle, which is unstable. Thus, after requiring that the solutions blow up, there will be no solutions with initial conditions off of the stable manifold.



Shifting back into the original coordinates, we draw the same conclusions with $(-4,-9/2)$.



We can see that, after requiring $C_1 = 0$, if we want to satisfy general initial condition $(x_0,y_0)$, we need $$C_2 = x_0 + 4 \ C_2 = frac{1}{2}y_0 + frac{9}{4}$$



which means that only initial conditions on the line $y = 2x+frac{7}{2}$ will lead to stable solutions, i.e. this line is the stable manifold.



We can clearly see that $0neq 2(0.5)+frac{7}{2}$, so the given initial conditions will not lead to a stable solution.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 0:23

























answered Dec 16 '18 at 0:08









AlexanderJ93AlexanderJ93

6,137823




6,137823












  • $begingroup$
    So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:31












  • $begingroup$
    Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
    $endgroup$
    – AlexanderJ93
    Dec 16 '18 at 0:34


















  • $begingroup$
    So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:31












  • $begingroup$
    Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
    $endgroup$
    – AlexanderJ93
    Dec 16 '18 at 0:34
















$begingroup$
So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
$endgroup$
– Justyen77
Dec 16 '18 at 0:31






$begingroup$
So, it would be pointless to check these initial conditions after requiring $C_1=0$ since the equations would not reach the stable manifold (different terminology for saddle path I assume) I think my understanding is now correct. Thanks!
$endgroup$
– Justyen77
Dec 16 '18 at 0:31














$begingroup$
Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
$endgroup$
– AlexanderJ93
Dec 16 '18 at 0:34




$begingroup$
Yeah, that's right. The stable and unstable manifolds are the two lines intersecting at the saddle point, on which "points" towards the equilibrium and one which "points" away.
$endgroup$
– AlexanderJ93
Dec 16 '18 at 0:34











0












$begingroup$

Hint :



It is $x(0) = 1/2$, thus :



$$x(0) = 0 Rightarrow 2C_1 + C_2 -4 = 1/2$$



Also, $y(0) = 0$, thus :



$$y(0) = 0 Rightarrow C_1 + 2C_2- 9/2 = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:24
















0












$begingroup$

Hint :



It is $x(0) = 1/2$, thus :



$$x(0) = 0 Rightarrow 2C_1 + C_2 -4 = 1/2$$



Also, $y(0) = 0$, thus :



$$y(0) = 0 Rightarrow C_1 + 2C_2- 9/2 = 0$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:24














0












0








0





$begingroup$

Hint :



It is $x(0) = 1/2$, thus :



$$x(0) = 0 Rightarrow 2C_1 + C_2 -4 = 1/2$$



Also, $y(0) = 0$, thus :



$$y(0) = 0 Rightarrow C_1 + 2C_2- 9/2 = 0$$






share|cite|improve this answer









$endgroup$



Hint :



It is $x(0) = 1/2$, thus :



$$x(0) = 0 Rightarrow 2C_1 + C_2 -4 = 1/2$$



Also, $y(0) = 0$, thus :



$$y(0) = 0 Rightarrow C_1 + 2C_2- 9/2 = 0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 0:06









RebellosRebellos

14.5k31246




14.5k31246












  • $begingroup$
    So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:24


















  • $begingroup$
    So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
    $endgroup$
    – Justyen77
    Dec 16 '18 at 0:24
















$begingroup$
So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
$endgroup$
– Justyen77
Dec 16 '18 at 0:24




$begingroup$
So $C_1=3/2$ and $C_2=1.5$ Just to be sure, is this intuition correct: To put it crudely: I made the mistake in that I used the initial conditions after setting $C_1 = 0$. And thanks!
$endgroup$
– Justyen77
Dec 16 '18 at 0:24


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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