Solve this: $frac{dx}{dt} = [a-(b+c)]x-ax^2$ by separation of variables












0












$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31


















0












$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31
















0












0








0





$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 12:39









Davide Giraudo

127k16151264




127k16151264










asked Dec 12 '18 at 12:11









countduckulacountduckula

1




1












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31




















  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31


















$begingroup$
"Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
$endgroup$
– Did
Dec 12 '18 at 14:46






$begingroup$
"Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
$endgroup$
– Did
Dec 12 '18 at 14:46














$begingroup$
True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
$endgroup$
– countduckula
Dec 12 '18 at 15:21




$begingroup$
True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
$endgroup$
– countduckula
Dec 12 '18 at 15:21












$begingroup$
$$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
$endgroup$
– Did
Dec 13 '18 at 22:31






$begingroup$
$$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
$endgroup$
– Did
Dec 13 '18 at 22:31












1 Answer
1






active

oldest

votes


















-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49











Your Answer





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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036608%2fsolve-this-fracdxdt-a-bcx-ax2-by-separation-of-variables%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









-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49
















-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49














-1












-1








-1





$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$



Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 14:43

























answered Dec 12 '18 at 12:27









TheD0ubleTTheD0ubleT

39218




39218












  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49


















  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49
















$begingroup$
There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
$endgroup$
– Satish Ramanathan
Dec 12 '18 at 12:53




$begingroup$
There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
$endgroup$
– Satish Ramanathan
Dec 12 '18 at 12:53












$begingroup$
I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
$endgroup$
– TheD0ubleT
Dec 12 '18 at 14:45




$begingroup$
I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
$endgroup$
– TheD0ubleT
Dec 12 '18 at 14:45












$begingroup$
Why hide carefully the core of the proof?
$endgroup$
– Did
Dec 12 '18 at 14:49




$begingroup$
Why hide carefully the core of the proof?
$endgroup$
– Did
Dec 12 '18 at 14:49


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036608%2fsolve-this-fracdxdt-a-bcx-ax2-by-separation-of-variables%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more

web3.py web3.isConnected() returns false always

Image
1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
Read more