Mixing
Solving a separable differential equation with limits
$begingroup$
Consider this equation, where $a$,$b$ and $k$ are constants,
$$frac{dy}{dt} = k(a-y)(b-y)$$
Assuming an initial condition $y=0$ at $t=0$
- Solve this equation for $y(t)$ for the case $a=b$.
- Solve this equation for the case $0 lt a lt b$
- By considering the limit $b rightarrow a$ in 2) show that the two results are consistent with each other.
So I am confident with my result for (i), through separating the variables, integrating then finding the constant of integration by using the initial conditions, resulting in:
$$y(t) = -frac{1}{kt}$$
However I am confused as to how to solve for the case of limits ?
Any help would be appreciated.
ordinary-differential-equations
edited Jan 31 at 10:28
Community♦
1
asked Jan 29 at 17:49
kaukerkauker
1
$endgroup$
add a comment |
$begingroup$
Consider this equation, where $a$,$b$ and $k$ are constants,
$$frac{dy}{dt} = k(a-y)(b-y)$$
Assuming an initial condition $y=0$ at $t=0$
- Solve this equation for $y(t)$ for the case $a=b$.
- Solve this equation for the case $0 lt a lt b$
- By considering the limit $b rightarrow a$ in 2) show that the two results are consistent with each other.
So I am confident with my result for (i), through separating the variables, integrating then finding the constant of integration by using the initial conditions, resulting in:
$$y(t) = -frac{1}{kt}$$
However I am confused as to how to solve for the case of limits ?
Any help would be appreciated.
ordinary-differential-equations
edited Jan 31 at 10:28
Community♦
1
asked Jan 29 at 17:49
kaukerkauker
1
$endgroup$
1
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
1
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49
add a comment |
$begingroup$
Consider this equation, where $a$,$b$ and $k$ are constants,
$$frac{dy}{dt} = k(a-y)(b-y)$$
Assuming an initial condition $y=0$ at $t=0$
- Solve this equation for $y(t)$ for the case $a=b$.
- Solve this equation for the case $0 lt a lt b$
- By considering the limit $b rightarrow a$ in 2) show that the two results are consistent with each other.
So I am confident with my result for (i), through separating the variables, integrating then finding the constant of integration by using the initial conditions, resulting in:
$$y(t) = -frac{1}{kt}$$
However I am confused as to how to solve for the case of limits ?
Any help would be appreciated.
ordinary-differential-equations
edited Jan 31 at 10:28
Community♦
1
asked Jan 29 at 17:49
kaukerkauker
1
$endgroup$
Consider this equation, where $a$,$b$ and $k$ are constants,
$$frac{dy}{dt} = k(a-y)(b-y)$$
Assuming an initial condition $y=0$ at $t=0$
- Solve this equation for $y(t)$ for the case $a=b$.
- Solve this equation for the case $0 lt a lt b$
- By considering the limit $b rightarrow a$ in 2) show that the two results are consistent with each other.
So I am confident with my result for (i), through separating the variables, integrating then finding the constant of integration by using the initial conditions, resulting in:
$$y(t) = -frac{1}{kt}$$
However I am confused as to how to solve for the case of limits ?
Any help would be appreciated.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 31 at 10:28
Community♦
1
asked Jan 29 at 17:49
kaukerkauker
1
edited Jan 31 at 10:28
Community♦
1
asked Jan 29 at 17:49
kaukerkauker
1
edited Jan 31 at 10:28
Community♦
1
edited Jan 31 at 10:28
Community♦
1
edited Jan 31 at 10:28
Community♦
1
1
asked Jan 29 at 17:49
kaukerkauker
1
asked Jan 29 at 17:49
kaukerkauker
1
asked Jan 29 at 17:49
kaukerkauker
1
1
1
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
1
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49
add a comment |
1
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
1
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49
1
1
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
1
1
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer is not correct. If you plug $y = -frac{1}{kt}$ into the ODE you get
$$ frac{dy}{dt} = frac{1}{kt^2} = ky^2 ne k(a-y)^2 $$
Furthermore, the initial condition $y(0)=0$ isn't satisfied.
I'll add more to this, if you edit the post to show your own working.
answered Jan 31 at 11:09
DylanDylan
14.2k31127
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092501%2fsolving-a-separable-differential-equation-with-limits%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer is not correct. If you plug $y = -frac{1}{kt}$ into the ODE you get
$$ frac{dy}{dt} = frac{1}{kt^2} = ky^2 ne k(a-y)^2 $$
Furthermore, the initial condition $y(0)=0$ isn't satisfied.
I'll add more to this, if you edit the post to show your own working.
answered Jan 31 at 11:09
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Your answer is not correct. If you plug $y = -frac{1}{kt}$ into the ODE you get
$$ frac{dy}{dt} = frac{1}{kt^2} = ky^2 ne k(a-y)^2 $$
Furthermore, the initial condition $y(0)=0$ isn't satisfied.
I'll add more to this, if you edit the post to show your own working.
answered Jan 31 at 11:09
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Your answer is not correct. If you plug $y = -frac{1}{kt}$ into the ODE you get
$$ frac{dy}{dt} = frac{1}{kt^2} = ky^2 ne k(a-y)^2 $$
Furthermore, the initial condition $y(0)=0$ isn't satisfied.
I'll add more to this, if you edit the post to show your own working.
answered Jan 31 at 11:09
DylanDylan
14.2k31127
$endgroup$
Your answer is not correct. If you plug $y = -frac{1}{kt}$ into the ODE you get
$$ frac{dy}{dt} = frac{1}{kt^2} = ky^2 ne k(a-y)^2 $$
Furthermore, the initial condition $y(0)=0$ isn't satisfied.
I'll add more to this, if you edit the post to show your own working.
answered Jan 31 at 11:09
DylanDylan
14.2k31127
answered Jan 31 at 11:09
DylanDylan
14.2k31127
answered Jan 31 at 11:09
DylanDylan
14.2k31127
answered Jan 31 at 11:09
DylanDylan
14.2k31127
14.2k31127
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3092501%2fsolving-a-separable-differential-equation-with-limits%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Are you sure you got the right solution for the first point? If $a=b$ then the equation is $dy/dx = k(a-y)^2$ and by separation of variables I get $y(x) = a^2kx/(akx+1)$.
$endgroup$
– Gibbs
Jan 29 at 18:04
$begingroup$
For $b neq a$ you can use the partial fraction decomposition $displaystyle frac{1}{(a-y)(b-y)} = frac{1}{b-a} left( frac{1}{a-y} - frac{1}{b-y} right)$. After finding the general solution take the limit as $b rightarrow a$.
$endgroup$
– Christoph
Jan 30 at 2:52
$begingroup$
@Gibbs apologies, the post had been edited, it is dy/dt not dy/dx
$endgroup$
– kauker
Jan 31 at 10:02
$begingroup$
The name of the variable does not matter. Actually now I read $dy/dx$ in the first equation and $t=0$ as initial data. There is no consistency in the notation. Anyway, I suggest you try again to compute the solution in the first case.
$endgroup$
– Gibbs
Jan 31 at 10:06
1
$begingroup$
Something is definitely wrong here. The solution should depend on $a$, and your solution is not $0$ at $t=0$. Please show your work so we can identify the problem.
$endgroup$
– Dylan
Jan 31 at 10:49