Consider the autonomus equation $frac{dy}{dt}=-2(y-1)(y-2)(y-a)^2$
$begingroup$
Consider the autonomus equation $frac{dy}{dt}=-2(y-1)(y-2)(y-a)^2$, where $a$ is any real number.
Then,
$(1)$ Plot the phase diagram showing the solution curves.
$(2)$ show that new solution can be generated from the old solutions (in $(a)$) by time shifting i.e, replacing $y(t)$ by $y(t-t_0)$.
Answer:
$(a)$ I have drawn the phase plot showing the solutions.
Please help me with the part $(b)$.
If we replace $y(t)$ by $y(t-t_0)$, then we have
$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$.
How to confirm that we get a new solution?
Help me
$
ordinary-differential-equations
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
$endgroup$
add a comment |
$begingroup$
Consider the autonomus equation $frac{dy}{dt}=-2(y-1)(y-2)(y-a)^2$, where $a$ is any real number.
Then,
$(1)$ Plot the phase diagram showing the solution curves.
$(2)$ show that new solution can be generated from the old solutions (in $(a)$) by time shifting i.e, replacing $y(t)$ by $y(t-t_0)$.
Answer:
$(a)$ I have drawn the phase plot showing the solutions.
Please help me with the part $(b)$.
If we replace $y(t)$ by $y(t-t_0)$, then we have
$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$.
How to confirm that we get a new solution?
Help me
$
ordinary-differential-equations
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
$endgroup$
add a comment |
$begingroup$
Consider the autonomus equation $frac{dy}{dt}=-2(y-1)(y-2)(y-a)^2$, where $a$ is any real number.
Then,
$(1)$ Plot the phase diagram showing the solution curves.
$(2)$ show that new solution can be generated from the old solutions (in $(a)$) by time shifting i.e, replacing $y(t)$ by $y(t-t_0)$.
Answer:
$(a)$ I have drawn the phase plot showing the solutions.
Please help me with the part $(b)$.
If we replace $y(t)$ by $y(t-t_0)$, then we have
$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$.
How to confirm that we get a new solution?
Help me
$
ordinary-differential-equations
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
$endgroup$
Consider the autonomus equation $frac{dy}{dt}=-2(y-1)(y-2)(y-a)^2$, where $a$ is any real number.
Then,
$(1)$ Plot the phase diagram showing the solution curves.
$(2)$ show that new solution can be generated from the old solutions (in $(a)$) by time shifting i.e, replacing $y(t)$ by $y(t-t_0)$.
Answer:
$(a)$ I have drawn the phase plot showing the solutions.
Please help me with the part $(b)$.
If we replace $y(t)$ by $y(t-t_0)$, then we have
$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$.
How to confirm that we get a new solution?
Help me
$
ordinary-differential-equations
ordinary-differential-equations
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
asked Jan 14 at 10:38
M. A. SARKARM. A. SARKAR
2,5021820
2,5021820
add a comment |
add a comment |
1 Answer
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$begingroup$
$$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$$
Change of variable : $quadtheta=t-t_0$ :
$$frac{dy(theta)}{dt}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
$dtheta=dt$
$$frac{dy(theta)}{dtheta}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
Thus $y(theta)$ is solution of the ODE.
Since $y(theta)=y(t-t_0)$ this proves that $y(t-t_0)$ is also solution.
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
add a comment |
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$begingroup$
$$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$$
Change of variable : $quadtheta=t-t_0$ :
$$frac{dy(theta)}{dt}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
$dtheta=dt$
$$frac{dy(theta)}{dtheta}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
Thus $y(theta)$ is solution of the ODE.
Since $y(theta)=y(t-t_0)$ this proves that $y(t-t_0)$ is also solution.
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
add a comment |
$begingroup$
$$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$$
Change of variable : $quadtheta=t-t_0$ :
$$frac{dy(theta)}{dt}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
$dtheta=dt$
$$frac{dy(theta)}{dtheta}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
Thus $y(theta)$ is solution of the ODE.
Since $y(theta)=y(t-t_0)$ this proves that $y(t-t_0)$ is also solution.
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
add a comment |
$begingroup$
$$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$$
Change of variable : $quadtheta=t-t_0$ :
$$frac{dy(theta)}{dt}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
$dtheta=dt$
$$frac{dy(theta)}{dtheta}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
Thus $y(theta)$ is solution of the ODE.
Since $y(theta)=y(t-t_0)$ this proves that $y(t-t_0)$ is also solution.
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
$endgroup$
$$frac{dy(t-t_0)}{dt}=-2(y(t-t_0)-1)(y(t-t_0)-2)(y(t-t_0)-a)^2$$
Change of variable : $quadtheta=t-t_0$ :
$$frac{dy(theta)}{dt}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
$dtheta=dt$
$$frac{dy(theta)}{dtheta}=-2(y(theta)-1)(y(theta)-2)(y(theta)-a)^2.$$
Thus $y(theta)$ is solution of the ODE.
Since $y(theta)=y(t-t_0)$ this proves that $y(t-t_0)$ is also solution.
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
answered Jan 16 at 18:59
JJacquelinJJacquelin
45.8k21858
45.8k21858
add a comment |
add a comment |
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