Mixing
Why can I state $ left| 3-y right|=e^{ -4t }$ iff $3-y=e^{ -4t } $?
$begingroup$
I have this differential equation problem.
$$ frac { dy }{ dt } =quad 12-4y,quad yleft( 0 right) quad =quad 0 $$
Walking through my steps.
$$ frac { dy }{ dt } =quad -4left( y-3 right) \ frac { 1 }{ left( y-3 right) } dyquad =quad -4dt\ int { frac { 1 }{ left( y-3 right) } dy } quad =quad -int { 4dt } \ ln { left( left| 3-y right| right) } quad =quad -4t+c\ { e }^{ ln { left( left| 3-y right| right) } }=quad { e }^{ -4t }+c\ left| 3-y right| quad =quad { e }^{ -4t }+c $$
Now my teacher has stated that we can assume the LHS is always going to be positive. She attempted to explain this to me but I was unable to understand the proof/logic behind stating the following two steps
$$
left| 3-y right| quad =quad { e }^{ -4t } +c\ quad 3-yquad =quad { e }^{ -4t }+c
$$
How can I show that $$ { e }^{ -4t } $$ is always positive?
ordinary-differential-equations intuition absolute-value
edited Jan 20 at 17:14
Did
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
$endgroup$
add a comment |
$begingroup$
I have this differential equation problem.
$$ frac { dy }{ dt } =quad 12-4y,quad yleft( 0 right) quad =quad 0 $$
Walking through my steps.
$$ frac { dy }{ dt } =quad -4left( y-3 right) \ frac { 1 }{ left( y-3 right) } dyquad =quad -4dt\ int { frac { 1 }{ left( y-3 right) } dy } quad =quad -int { 4dt } \ ln { left( left| 3-y right| right) } quad =quad -4t+c\ { e }^{ ln { left( left| 3-y right| right) } }=quad { e }^{ -4t }+c\ left| 3-y right| quad =quad { e }^{ -4t }+c $$
Now my teacher has stated that we can assume the LHS is always going to be positive. She attempted to explain this to me but I was unable to understand the proof/logic behind stating the following two steps
$$
left| 3-y right| quad =quad { e }^{ -4t } +c\ quad 3-yquad =quad { e }^{ -4t }+c
$$
How can I show that $$ { e }^{ -4t } $$ is always positive?
ordinary-differential-equations intuition absolute-value
edited Jan 20 at 17:14
Did
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
$endgroup$
$begingroup$
Right. That is my question. How can I show, that?
$endgroup$
– user1787331
Dec 11 '16 at 1:59
1
$begingroup$
The equivalence is false since it is possible that $3-y=-e^{-4t}$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:03
$begingroup$
There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
$endgroup$
– mrob
Dec 11 '16 at 2:06
$begingroup$
You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
$endgroup$
– Sangchul Lee
Dec 11 '16 at 2:12
$begingroup$
There are several mistakes still in your work. I will try to make my answer below more inclusive.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:24
add a comment |
$begingroup$
I have this differential equation problem.
$$ frac { dy }{ dt } =quad 12-4y,quad yleft( 0 right) quad =quad 0 $$
Walking through my steps.
$$ frac { dy }{ dt } =quad -4left( y-3 right) \ frac { 1 }{ left( y-3 right) } dyquad =quad -4dt\ int { frac { 1 }{ left( y-3 right) } dy } quad =quad -int { 4dt } \ ln { left( left| 3-y right| right) } quad =quad -4t+c\ { e }^{ ln { left( left| 3-y right| right) } }=quad { e }^{ -4t }+c\ left| 3-y right| quad =quad { e }^{ -4t }+c $$
Now my teacher has stated that we can assume the LHS is always going to be positive. She attempted to explain this to me but I was unable to understand the proof/logic behind stating the following two steps
$$
left| 3-y right| quad =quad { e }^{ -4t } +c\ quad 3-yquad =quad { e }^{ -4t }+c
$$
How can I show that $$ { e }^{ -4t } $$ is always positive?
ordinary-differential-equations intuition absolute-value
edited Jan 20 at 17:14
Did
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
$endgroup$
I have this differential equation problem.
$$ frac { dy }{ dt } =quad 12-4y,quad yleft( 0 right) quad =quad 0 $$
Walking through my steps.
$$ frac { dy }{ dt } =quad -4left( y-3 right) \ frac { 1 }{ left( y-3 right) } dyquad =quad -4dt\ int { frac { 1 }{ left( y-3 right) } dy } quad =quad -int { 4dt } \ ln { left( left| 3-y right| right) } quad =quad -4t+c\ { e }^{ ln { left( left| 3-y right| right) } }=quad { e }^{ -4t }+c\ left| 3-y right| quad =quad { e }^{ -4t }+c $$
Now my teacher has stated that we can assume the LHS is always going to be positive. She attempted to explain this to me but I was unable to understand the proof/logic behind stating the following two steps
$$
left| 3-y right| quad =quad { e }^{ -4t } +c\ quad 3-yquad =quad { e }^{ -4t }+c
$$
How can I show that $$ { e }^{ -4t } $$ is always positive?
ordinary-differential-equations intuition absolute-value
ordinary-differential-equations intuition absolute-value
edited Jan 20 at 17:14
Did
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
edited Jan 20 at 17:14
Did
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
edited Jan 20 at 17:14
Did
248k23225463
edited Jan 20 at 17:14
Did
248k23225463
edited Jan 20 at 17:14
Did
248k23225463
248k23225463
asked Dec 11 '16 at 1:55
user1787331user1787331
706
asked Dec 11 '16 at 1:55
user1787331user1787331
706
asked Dec 11 '16 at 1:55
user1787331user1787331
706
706
$begingroup$
Right. That is my question. How can I show, that?
$endgroup$
– user1787331
Dec 11 '16 at 1:59
1
$begingroup$
The equivalence is false since it is possible that $3-y=-e^{-4t}$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:03
$begingroup$
There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
$endgroup$
– mrob
Dec 11 '16 at 2:06
$begingroup$
You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
$endgroup$
– Sangchul Lee
Dec 11 '16 at 2:12
$begingroup$
There are several mistakes still in your work. I will try to make my answer below more inclusive.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:24
add a comment |
$begingroup$
Right. That is my question. How can I show, that?
$endgroup$
– user1787331
Dec 11 '16 at 1:59
1
$begingroup$
The equivalence is false since it is possible that $3-y=-e^{-4t}$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:03
$begingroup$
There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
$endgroup$
– mrob
Dec 11 '16 at 2:06
$begingroup$
You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
$endgroup$
– Sangchul Lee
Dec 11 '16 at 2:12
$begingroup$
There are several mistakes still in your work. I will try to make my answer below more inclusive.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:24
$begingroup$
Right. That is my question. How can I show, that?
$endgroup$
– user1787331
Dec 11 '16 at 1:59
$begingroup$
Right. That is my question. How can I show, that?
$endgroup$
– user1787331
Dec 11 '16 at 1:59
1
1
$begingroup$
The equivalence is false since it is possible that $3-y=-e^{-4t}$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:03
$begingroup$
The equivalence is false since it is possible that $3-y=-e^{-4t}$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:03
$begingroup$
There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
$endgroup$
– mrob
Dec 11 '16 at 2:06
$begingroup$
There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
$endgroup$
– mrob
Dec 11 '16 at 2:06
$begingroup$
You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
$endgroup$
– Sangchul Lee
Dec 11 '16 at 2:12
$begingroup$
You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
$endgroup$
– Sangchul Lee
Dec 11 '16 at 2:12
$begingroup$
There are several mistakes still in your work. I will try to make my answer below more inclusive.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:24
$begingroup$
There are several mistakes still in your work. I will try to make my answer below more inclusive.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given $dfrac{dy}{dt}=12-4y=4(3-y),quad y(0)=0$.
$$intdfrac{1}{3-y},dy=int4,dt$$
$$-lnvert 3-yvert=4t-c_0$$
$$ lnvert 3-yvert=c_0-4t $$
$$ vert3-yvert=e^{c_0}e^{-4t}=c_1e^{-4t} $$
$$3-y = pm c_1{ e }^{ -4t }=ce^{-4t}$$
$$ y=3-ce^{-4t}$$
is the solution
$y(0)=3-ccdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
add a comment |
$begingroup$
When you integrate $-4dt$ you should get
begin{equation}
-ln(|3-y|) = -4t + A_1
end{equation}
where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = pm A_2 e^{-4t}$ and we can just absorb $pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.
EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
$endgroup$
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Given $dfrac{dy}{dt}=12-4y=4(3-y),quad y(0)=0$.
$$intdfrac{1}{3-y},dy=int4,dt$$
$$-lnvert 3-yvert=4t-c_0$$
$$ lnvert 3-yvert=c_0-4t $$
$$ vert3-yvert=e^{c_0}e^{-4t}=c_1e^{-4t} $$
$$3-y = pm c_1{ e }^{ -4t }=ce^{-4t}$$
$$ y=3-ce^{-4t}$$
is the solution
$y(0)=3-ccdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
add a comment |
$begingroup$
Given $dfrac{dy}{dt}=12-4y=4(3-y),quad y(0)=0$.
$$intdfrac{1}{3-y},dy=int4,dt$$
$$-lnvert 3-yvert=4t-c_0$$
$$ lnvert 3-yvert=c_0-4t $$
$$ vert3-yvert=e^{c_0}e^{-4t}=c_1e^{-4t} $$
$$3-y = pm c_1{ e }^{ -4t }=ce^{-4t}$$
$$ y=3-ce^{-4t}$$
is the solution
$y(0)=3-ccdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
add a comment |
$begingroup$
Given $dfrac{dy}{dt}=12-4y=4(3-y),quad y(0)=0$.
$$intdfrac{1}{3-y},dy=int4,dt$$
$$-lnvert 3-yvert=4t-c_0$$
$$ lnvert 3-yvert=c_0-4t $$
$$ vert3-yvert=e^{c_0}e^{-4t}=c_1e^{-4t} $$
$$3-y = pm c_1{ e }^{ -4t }=ce^{-4t}$$
$$ y=3-ce^{-4t}$$
is the solution
$y(0)=3-ccdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
$endgroup$
Given $dfrac{dy}{dt}=12-4y=4(3-y),quad y(0)=0$.
$$intdfrac{1}{3-y},dy=int4,dt$$
$$-lnvert 3-yvert=4t-c_0$$
$$ lnvert 3-yvert=c_0-4t $$
$$ vert3-yvert=e^{c_0}e^{-4t}=c_1e^{-4t} $$
$$3-y = pm c_1{ e }^{ -4t }=ce^{-4t}$$
$$ y=3-ce^{-4t}$$
is the solution
$y(0)=3-ccdot1=0$ so $c=3$. Therefore $y(t)=3-3e^{-4t}$.
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
edited Dec 11 '16 at 2:59
Mathlover
3,6181022
3,6181022
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
answered Dec 11 '16 at 2:12
John Wayland BalesJohn Wayland Bales
14.5k21238
14.5k21238
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
add a comment |
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
Just factored out a -4, you can multiply it back in to get the same expression.
$endgroup$
– user1787331
Dec 11 '16 at 2:20
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
No, when you multiply it out you get $-12+4y$ not $12-4y$.
$endgroup$
– John Wayland Bales
Dec 11 '16 at 2:22
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
$begingroup$
sorry those were errors. in my latex
$endgroup$
– user1787331
Dec 11 '16 at 2:26
add a comment |
$begingroup$
When you integrate $-4dt$ you should get
begin{equation}
-ln(|3-y|) = -4t + A_1
end{equation}
where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = pm A_2 e^{-4t}$ and we can just absorb $pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.
EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
$endgroup$
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
add a comment |
$begingroup$
When you integrate $-4dt$ you should get
begin{equation}
-ln(|3-y|) = -4t + A_1
end{equation}
where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = pm A_2 e^{-4t}$ and we can just absorb $pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.
EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
$endgroup$
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
add a comment |
$begingroup$
When you integrate $-4dt$ you should get
begin{equation}
-ln(|3-y|) = -4t + A_1
end{equation}
where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = pm A_2 e^{-4t}$ and we can just absorb $pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.
EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
$endgroup$
When you integrate $-4dt$ you should get
begin{equation}
-ln(|3-y|) = -4t + A_1
end{equation}
where $C$ is an unknown constant. So your solution should be $|3 - y| = A_2e^{-4t}$ where $A_2 = e^{A_1}$. Therefore $3 - y = pm A_2 e^{-4t}$ and we can just absorb $pm$ into the constant $A_2$ and write $y = 3 + Ae^{-4t}$. The conclusion is you should get more than one solution for your ODE. Why is that? Because an ODE doesn't give a unique solution unless you have an initial condition $y(0) = $ something. Once an initial condition is known you can solve for $A$.
EDIT: Since $y(0) = 0$ we should have $A = -3$ ie. $y = 3 - 3e^{-4t}$.
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
edited Dec 11 '16 at 2:10
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
answered Dec 11 '16 at 2:05
user113988user113988
1,210818
1,210818
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
add a comment |
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
$begingroup$
updated my question. Thanks for your feedback.
$endgroup$
– user1787331
Dec 11 '16 at 2:08
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Right. That is my question. How can I show, that?
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– user1787331
Dec 11 '16 at 1:59
1
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The equivalence is false since it is possible that $3-y=-e^{-4t}$.
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– John Wayland Bales
Dec 11 '16 at 2:03
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There are a lot of errors in your work. Maybe they have something to do with your issue. The problems are in factoring out $-4$ and simplifying $e^{-ln|3-y|}$.
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– mrob
Dec 11 '16 at 2:06
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You missed the constant of integration. This gives $3 - y(t) = Ae^{-4t} $$ for some $A neq 0$.
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– Sangchul Lee
Dec 11 '16 at 2:12
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There are several mistakes still in your work. I will try to make my answer below more inclusive.
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– John Wayland Bales
Dec 11 '16 at 2:24