write a differential equation $(frac{dy}{dt} = ay+b)$ whose solutions have the required behavior as $...
$begingroup$
Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).
I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?
ordinary-differential-equations
edited Jan 21 at 5:59
user549397
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
$endgroup$
add a comment |
$begingroup$
Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).
I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?
ordinary-differential-equations
edited Jan 21 at 5:59
user549397
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
$endgroup$
add a comment |
$begingroup$
Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).
I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?
ordinary-differential-equations
edited Jan 21 at 5:59
user549397
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
$endgroup$
Let all other solutions diverge from $y = frac{3}{4}$. write a differential equation $(frac{dy}{dt} = ay +b)$ whose solutions have the required behavior as t goes to infinity. Explain how you came up with the differential equation).
I think the answer is $y' = 4y - 3$ but I am not sure. Can someone help me produce the answer with complete working?
ordinary-differential-equations
ordinary-differential-equations
edited Jan 21 at 5:59
user549397
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
edited Jan 21 at 5:59
user549397
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
edited Jan 21 at 5:59
user549397
1,5101418
edited Jan 21 at 5:59
user549397
1,5101418
edited Jan 21 at 5:59
user549397
1,5101418
1,5101418
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
asked Sep 22 '16 at 12:07
AnonymousAnonymous
214
214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
well i think you are really close to the solution:
$$y'=-4y-3$$
Homogeneous part would be:
$$y'+ 4y = 0$$
which leads to:
$$y0=Ce^{-4t}$$
particular part would be (with initial assumption $yp=ax+b$):
$$yp=3/4$$
which leads to:
$$y(t)=Ce^{-4t}+3/4$$
$$limlimits_{t to infty} y(t)=3/4$$
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
$endgroup$
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
add a comment |
$begingroup$
Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that
If $a > 0$ then $y^*$ is unstable.
If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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active
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votes
$begingroup$
well i think you are really close to the solution:
$$y'=-4y-3$$
Homogeneous part would be:
$$y'+ 4y = 0$$
which leads to:
$$y0=Ce^{-4t}$$
particular part would be (with initial assumption $yp=ax+b$):
$$yp=3/4$$
which leads to:
$$y(t)=Ce^{-4t}+3/4$$
$$limlimits_{t to infty} y(t)=3/4$$
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
$endgroup$
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
add a comment |
$begingroup$
well i think you are really close to the solution:
$$y'=-4y-3$$
Homogeneous part would be:
$$y'+ 4y = 0$$
which leads to:
$$y0=Ce^{-4t}$$
particular part would be (with initial assumption $yp=ax+b$):
$$yp=3/4$$
which leads to:
$$y(t)=Ce^{-4t}+3/4$$
$$limlimits_{t to infty} y(t)=3/4$$
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
$endgroup$
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
add a comment |
$begingroup$
well i think you are really close to the solution:
$$y'=-4y-3$$
Homogeneous part would be:
$$y'+ 4y = 0$$
which leads to:
$$y0=Ce^{-4t}$$
particular part would be (with initial assumption $yp=ax+b$):
$$yp=3/4$$
which leads to:
$$y(t)=Ce^{-4t}+3/4$$
$$limlimits_{t to infty} y(t)=3/4$$
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
$endgroup$
well i think you are really close to the solution:
$$y'=-4y-3$$
Homogeneous part would be:
$$y'+ 4y = 0$$
which leads to:
$$y0=Ce^{-4t}$$
particular part would be (with initial assumption $yp=ax+b$):
$$yp=3/4$$
which leads to:
$$y(t)=Ce^{-4t}+3/4$$
$$limlimits_{t to infty} y(t)=3/4$$
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
edited Sep 22 '16 at 13:00
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
answered Sep 22 '16 at 12:46
arash javanarash javan
1397
1397
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
add a comment |
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
Is it y' = -4y - 3 or y' = 4y - 3
$endgroup$
– Anonymous
Sep 22 '16 at 13:03
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
y'= -4y-3 try to solve some similar equations.
$endgroup$
– arash javan
Sep 22 '16 at 13:07
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
$begingroup$
@Anonymous mark the answer plz, if it was your solution
$endgroup$
– arash javan
Sep 23 '16 at 14:09
add a comment |
$begingroup$
Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that
If $a > 0$ then $y^*$ is unstable.
If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
$endgroup$
add a comment |
$begingroup$
Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that
If $a > 0$ then $y^*$ is unstable.
If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
$endgroup$
add a comment |
$begingroup$
Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that
If $a > 0$ then $y^*$ is unstable.
If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
$endgroup$
Differential equation $y' = ax +b = f(y)$ is a nonautonuous equation. If a = 0, this equation becomes $y' = b$. Otherwise, if $a ne 0$, the equation has a unique equibrium point $y^* = -b/a$ and $f'(y^*) = a$. Therefore, by Lyapunov indirect method, we conclude that
If $a > 0$ then $y^*$ is unstable.
If $a < 0$ then $y^*$ is globally asymptotically stable. It implies that all solutions converge to $y^*$ as $t to infty$.
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
answered Dec 17 '18 at 19:12
MichaelCarrickMichaelCarrick
1097
1097
add a comment |
add a comment |
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