Can I add the constant of integration when and wherever I want?












0












$begingroup$


Lets consider this simple differential equation:
$dy/dx=beta y ; y(0)=2$



The solution is: $ln(y)=beta x$



From this points I have two options, adding $C$ and then find $y$ or first find $y$ and then adding $C$:



a) $ln(y)=beta x+CRightarrow y=e^{beta x+C} Rightarrow y=e^{C} e^{beta x} Rightarrow mathbf{y=Ce^{beta x}}$



b) $mathbf{y=e^{beta x}+C}$



And with boundary conditions:



a) $mathbf{y=2e^{beta x}}$



b) $mathbf{y=e^{beta x}+1}$



(a) and (b) are different functions, so is it correct/not correct to add the constant whenever I want? Am I missing something here?



Thanks in advance






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $lan(y)$ mean $ln(y)$?
    $endgroup$
    – littleO
    Dec 16 '18 at 7:33










  • $begingroup$
    Yes, I've edited
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:05
















0












$begingroup$


Lets consider this simple differential equation:
$dy/dx=beta y ; y(0)=2$



The solution is: $ln(y)=beta x$



From this points I have two options, adding $C$ and then find $y$ or first find $y$ and then adding $C$:



a) $ln(y)=beta x+CRightarrow y=e^{beta x+C} Rightarrow y=e^{C} e^{beta x} Rightarrow mathbf{y=Ce^{beta x}}$



b) $mathbf{y=e^{beta x}+C}$



And with boundary conditions:



a) $mathbf{y=2e^{beta x}}$



b) $mathbf{y=e^{beta x}+1}$



(a) and (b) are different functions, so is it correct/not correct to add the constant whenever I want? Am I missing something here?



Thanks in advance






integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does $lan(y)$ mean $ln(y)$?
    $endgroup$
    – littleO
    Dec 16 '18 at 7:33










  • $begingroup$
    Yes, I've edited
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:05














0












0








0





$begingroup$


Lets consider this simple differential equation:
$dy/dx=beta y ; y(0)=2$



The solution is: $ln(y)=beta x$



From this points I have two options, adding $C$ and then find $y$ or first find $y$ and then adding $C$:



a) $ln(y)=beta x+CRightarrow y=e^{beta x+C} Rightarrow y=e^{C} e^{beta x} Rightarrow mathbf{y=Ce^{beta x}}$



b) $mathbf{y=e^{beta x}+C}$



And with boundary conditions:



a) $mathbf{y=2e^{beta x}}$



b) $mathbf{y=e^{beta x}+1}$



(a) and (b) are different functions, so is it correct/not correct to add the constant whenever I want? Am I missing something here?



Thanks in advance






integration







share|cite|improve this question











$endgroup$




Lets consider this simple differential equation:
$dy/dx=beta y ; y(0)=2$



The solution is: $ln(y)=beta x$



From this points I have two options, adding $C$ and then find $y$ or first find $y$ and then adding $C$:



a) $ln(y)=beta x+CRightarrow y=e^{beta x+C} Rightarrow y=e^{C} e^{beta x} Rightarrow mathbf{y=Ce^{beta x}}$



b) $mathbf{y=e^{beta x}+C}$



And with boundary conditions:



a) $mathbf{y=2e^{beta x}}$



b) $mathbf{y=e^{beta x}+1}$



(a) and (b) are different functions, so is it correct/not correct to add the constant whenever I want? Am I missing something here?



Thanks in advance






integration




integration






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 8:02







dor gotleyb

















asked Dec 16 '18 at 7:29









dor gotleybdor gotleyb

155




155












  • $begingroup$
    Does $lan(y)$ mean $ln(y)$?
    $endgroup$
    – littleO
    Dec 16 '18 at 7:33










  • $begingroup$
    Yes, I've edited
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:05


















  • $begingroup$
    Does $lan(y)$ mean $ln(y)$?
    $endgroup$
    – littleO
    Dec 16 '18 at 7:33










  • $begingroup$
    Yes, I've edited
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:05
















$begingroup$
Does $lan(y)$ mean $ln(y)$?
$endgroup$
– littleO
Dec 16 '18 at 7:33




$begingroup$
Does $lan(y)$ mean $ln(y)$?
$endgroup$
– littleO
Dec 16 '18 at 7:33












$begingroup$
Yes, I've edited
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:05




$begingroup$
Yes, I've edited
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:05










2 Answers
2






active

oldest

votes


















0












$begingroup$

The constant of integration, $C$, must go in front, as in $y = Ce^{beta t}$, and not added, as in $y = e^{beta t} + C$. This is because, if $y = e^{beta t} + C$, then, $y' = beta e^{beta t} ne beta*y = beta*(e^{beta t} + C)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11



















1












$begingroup$

I think the derivation in a) will be more clear if we include a step that has been omitted. We first note that
$$
frac{y'(x)}{y(x)} = beta
$$
(assuming that $y(x)$ is not zero).
Then we take antiderivatives of both sides, concluding that
$$
ln(y(x)) = beta x + C
$$
for some constant $C$. (Here we are also assuming that $y(x)$ is always positive.)



It follows that
$$
y(x) = e^{beta x + C} = K e^{beta x}
$$
(where $K = e^C$).



We can now use the initial condition $y(0) = 2$ to conclude that
$y(x) = 2 e^{beta x}$.



One can't just add a constant of integration wherever one wants. The fact is that the general antiderivative of a continuous function $f$ on an interval $I$ is $F(x) + C$, where $F$ is any particular antiderivative of $f$. Because of this fact, we must carefully to remember to include "$+C$" when taking antiderivatives (but not at other times when we are not taking antiderivatives).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11











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%2f3042336%2fcan-i-add-the-constant-of-integration-when-and-wherever-i-want%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The constant of integration, $C$, must go in front, as in $y = Ce^{beta t}$, and not added, as in $y = e^{beta t} + C$. This is because, if $y = e^{beta t} + C$, then, $y' = beta e^{beta t} ne beta*y = beta*(e^{beta t} + C)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11
















0












$begingroup$

The constant of integration, $C$, must go in front, as in $y = Ce^{beta t}$, and not added, as in $y = e^{beta t} + C$. This is because, if $y = e^{beta t} + C$, then, $y' = beta e^{beta t} ne beta*y = beta*(e^{beta t} + C)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11














0












0








0





$begingroup$

The constant of integration, $C$, must go in front, as in $y = Ce^{beta t}$, and not added, as in $y = e^{beta t} + C$. This is because, if $y = e^{beta t} + C$, then, $y' = beta e^{beta t} ne beta*y = beta*(e^{beta t} + C)$.






share|cite|improve this answer









$endgroup$



The constant of integration, $C$, must go in front, as in $y = Ce^{beta t}$, and not added, as in $y = e^{beta t} + C$. This is because, if $y = e^{beta t} + C$, then, $y' = beta e^{beta t} ne beta*y = beta*(e^{beta t} + C)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 7:33









D.B.D.B.

1,2438




1,2438












  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11


















  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11
















$begingroup$
Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:11




$begingroup$
Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:11











1












$begingroup$

I think the derivation in a) will be more clear if we include a step that has been omitted. We first note that
$$
frac{y'(x)}{y(x)} = beta
$$
(assuming that $y(x)$ is not zero).
Then we take antiderivatives of both sides, concluding that
$$
ln(y(x)) = beta x + C
$$
for some constant $C$. (Here we are also assuming that $y(x)$ is always positive.)



It follows that
$$
y(x) = e^{beta x + C} = K e^{beta x}
$$
(where $K = e^C$).



We can now use the initial condition $y(0) = 2$ to conclude that
$y(x) = 2 e^{beta x}$.



One can't just add a constant of integration wherever one wants. The fact is that the general antiderivative of a continuous function $f$ on an interval $I$ is $F(x) + C$, where $F$ is any particular antiderivative of $f$. Because of this fact, we must carefully to remember to include "$+C$" when taking antiderivatives (but not at other times when we are not taking antiderivatives).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11
















1












$begingroup$

I think the derivation in a) will be more clear if we include a step that has been omitted. We first note that
$$
frac{y'(x)}{y(x)} = beta
$$
(assuming that $y(x)$ is not zero).
Then we take antiderivatives of both sides, concluding that
$$
ln(y(x)) = beta x + C
$$
for some constant $C$. (Here we are also assuming that $y(x)$ is always positive.)



It follows that
$$
y(x) = e^{beta x + C} = K e^{beta x}
$$
(where $K = e^C$).



We can now use the initial condition $y(0) = 2$ to conclude that
$y(x) = 2 e^{beta x}$.



One can't just add a constant of integration wherever one wants. The fact is that the general antiderivative of a continuous function $f$ on an interval $I$ is $F(x) + C$, where $F$ is any particular antiderivative of $f$. Because of this fact, we must carefully to remember to include "$+C$" when taking antiderivatives (but not at other times when we are not taking antiderivatives).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11














1












1








1





$begingroup$

I think the derivation in a) will be more clear if we include a step that has been omitted. We first note that
$$
frac{y'(x)}{y(x)} = beta
$$
(assuming that $y(x)$ is not zero).
Then we take antiderivatives of both sides, concluding that
$$
ln(y(x)) = beta x + C
$$
for some constant $C$. (Here we are also assuming that $y(x)$ is always positive.)



It follows that
$$
y(x) = e^{beta x + C} = K e^{beta x}
$$
(where $K = e^C$).



We can now use the initial condition $y(0) = 2$ to conclude that
$y(x) = 2 e^{beta x}$.



One can't just add a constant of integration wherever one wants. The fact is that the general antiderivative of a continuous function $f$ on an interval $I$ is $F(x) + C$, where $F$ is any particular antiderivative of $f$. Because of this fact, we must carefully to remember to include "$+C$" when taking antiderivatives (but not at other times when we are not taking antiderivatives).






share|cite|improve this answer











$endgroup$



I think the derivation in a) will be more clear if we include a step that has been omitted. We first note that
$$
frac{y'(x)}{y(x)} = beta
$$
(assuming that $y(x)$ is not zero).
Then we take antiderivatives of both sides, concluding that
$$
ln(y(x)) = beta x + C
$$
for some constant $C$. (Here we are also assuming that $y(x)$ is always positive.)



It follows that
$$
y(x) = e^{beta x + C} = K e^{beta x}
$$
(where $K = e^C$).



We can now use the initial condition $y(0) = 2$ to conclude that
$y(x) = 2 e^{beta x}$.



One can't just add a constant of integration wherever one wants. The fact is that the general antiderivative of a continuous function $f$ on an interval $I$ is $F(x) + C$, where $F$ is any particular antiderivative of $f$. Because of this fact, we must carefully to remember to include "$+C$" when taking antiderivatives (but not at other times when we are not taking antiderivatives).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 7:48

























answered Dec 16 '18 at 7:42









littleOlittleO

29.8k646109




29.8k646109












  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11


















  • $begingroup$
    Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
    $endgroup$
    – dor gotleyb
    Dec 16 '18 at 8:11
















$begingroup$
Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:11




$begingroup$
Ok, thank you very much, I guess I had a misconception, its been long time since I studied it.
$endgroup$
– dor gotleyb
Dec 16 '18 at 8:11


















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%2f3042336%2fcan-i-add-the-constant-of-integration-when-and-wherever-i-want%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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more