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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

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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.

1,26518




1,26518












  • $begingroup$
    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

29.9k647110




29.9k647110












  • $begingroup$
    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


















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