$dy/dx=e^{-y}$ differential equation
up vote
0
down vote
favorite
I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:
However that is not a correct solution. How can I derrive the true solution?
This is not correct as:
I forgot to write ln, the right side hasn't changed
I used TI-Nspire CX CAS student software to come to that conclusion:
Weird it's given me a wrong answer
differential-equations
asked Nov 28 at 16:21
Ryan Cameron
407
add a comment |
up vote
0
down vote
favorite
I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:
However that is not a correct solution. How can I derrive the true solution?
This is not correct as:
I forgot to write ln, the right side hasn't changed
I used TI-Nspire CX CAS student software to come to that conclusion:
Weird it's given me a wrong answer
differential-equations
asked Nov 28 at 16:21
Ryan Cameron
407
Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:
However that is not a correct solution. How can I derrive the true solution?
This is not correct as:
I forgot to write ln, the right side hasn't changed
I used TI-Nspire CX CAS student software to come to that conclusion:
Weird it's given me a wrong answer
differential-equations
asked Nov 28 at 16:21
Ryan Cameron
407
I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:
However that is not a correct solution. How can I derrive the true solution?
This is not correct as:
I forgot to write ln, the right side hasn't changed
I used TI-Nspire CX CAS student software to come to that conclusion:
Weird it's given me a wrong answer
differential-equations
differential-equations
asked Nov 28 at 16:21
Ryan Cameron
407
asked Nov 28 at 16:21
Ryan Cameron
407
edited Nov 28 at 16:48
asked Nov 28 at 16:21
Ryan Cameron
407
asked Nov 28 at 16:21
Ryan Cameron
407
asked Nov 28 at 16:21
Ryan Cameron
407
407
Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23
add a comment |
Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23
Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23
Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
This is not correct as:
That derivative is wrong and why does the $ln$ disappear in the exponent in the right-hand side?
If $y=ln(x+c)$, then $y'=tfrac{1}{x+c}$ but also:
$$e^{-y}=e^{-ln(x+c)}=frac{1}{e^{ln(x+c)}}=tfrac{1}{x+c}$$
The derivative of $ln x$ is $tfrac{1}{x}$ so then by the chain rule:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$
answered Nov 28 at 16:28
StackTD
22k1946
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
add a comment |
up vote
2
down vote
You can check this is actually the right solution
$$
frac{{rm d}y}{{rm d}x} = frac{1}{x + c} tag{1}
$$
and
$$
e^{-y} = frac{1}{e^y} = frac{1}{e^{ln(x + c)}} = frac{1}{x + c} tag{2}
$$
So yes, $y(x) = ln |x + c|$ is the solution
answered Nov 28 at 16:27
caverac
12.8k21028
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is not correct as:
That derivative is wrong and why does the $ln$ disappear in the exponent in the right-hand side?
If $y=ln(x+c)$, then $y'=tfrac{1}{x+c}$ but also:
$$e^{-y}=e^{-ln(x+c)}=frac{1}{e^{ln(x+c)}}=tfrac{1}{x+c}$$
The derivative of $ln x$ is $tfrac{1}{x}$ so then by the chain rule:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$
answered Nov 28 at 16:28
StackTD
22k1946
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
add a comment |
up vote
2
down vote
accepted
This is not correct as:
That derivative is wrong and why does the $ln$ disappear in the exponent in the right-hand side?
If $y=ln(x+c)$, then $y'=tfrac{1}{x+c}$ but also:
$$e^{-y}=e^{-ln(x+c)}=frac{1}{e^{ln(x+c)}}=tfrac{1}{x+c}$$
The derivative of $ln x$ is $tfrac{1}{x}$ so then by the chain rule:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$
answered Nov 28 at 16:28
StackTD
22k1946
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is not correct as:
That derivative is wrong and why does the $ln$ disappear in the exponent in the right-hand side?
If $y=ln(x+c)$, then $y'=tfrac{1}{x+c}$ but also:
$$e^{-y}=e^{-ln(x+c)}=frac{1}{e^{ln(x+c)}}=tfrac{1}{x+c}$$
The derivative of $ln x$ is $tfrac{1}{x}$ so then by the chain rule:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$
answered Nov 28 at 16:28
StackTD
22k1946
This is not correct as:
That derivative is wrong and why does the $ln$ disappear in the exponent in the right-hand side?
If $y=ln(x+c)$, then $y'=tfrac{1}{x+c}$ but also:
$$e^{-y}=e^{-ln(x+c)}=frac{1}{e^{ln(x+c)}}=tfrac{1}{x+c}$$
The derivative of $ln x$ is $tfrac{1}{x}$ so then by the chain rule:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$
answered Nov 28 at 16:28
StackTD
22k1946
edited Nov 28 at 16:50
answered Nov 28 at 16:28
StackTD
22k1946
answered Nov 28 at 16:28
StackTD
22k1946
answered Nov 28 at 16:28
StackTD
22k1946
22k1946
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
add a comment |
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 at 16:49
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 at 16:50
add a comment |
up vote
2
down vote
You can check this is actually the right solution
$$
frac{{rm d}y}{{rm d}x} = frac{1}{x + c} tag{1}
$$
and
$$
e^{-y} = frac{1}{e^y} = frac{1}{e^{ln(x + c)}} = frac{1}{x + c} tag{2}
$$
So yes, $y(x) = ln |x + c|$ is the solution
answered Nov 28 at 16:27
caverac
12.8k21028
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
add a comment |
up vote
2
down vote
You can check this is actually the right solution
$$
frac{{rm d}y}{{rm d}x} = frac{1}{x + c} tag{1}
$$
and
$$
e^{-y} = frac{1}{e^y} = frac{1}{e^{ln(x + c)}} = frac{1}{x + c} tag{2}
$$
So yes, $y(x) = ln |x + c|$ is the solution
answered Nov 28 at 16:27
caverac
12.8k21028
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
add a comment |
up vote
2
down vote
up vote
2
down vote
You can check this is actually the right solution
$$
frac{{rm d}y}{{rm d}x} = frac{1}{x + c} tag{1}
$$
and
$$
e^{-y} = frac{1}{e^y} = frac{1}{e^{ln(x + c)}} = frac{1}{x + c} tag{2}
$$
So yes, $y(x) = ln |x + c|$ is the solution
answered Nov 28 at 16:27
caverac
12.8k21028
You can check this is actually the right solution
$$
frac{{rm d}y}{{rm d}x} = frac{1}{x + c} tag{1}
$$
and
$$
e^{-y} = frac{1}{e^y} = frac{1}{e^{ln(x + c)}} = frac{1}{x + c} tag{2}
$$
So yes, $y(x) = ln |x + c|$ is the solution
answered Nov 28 at 16:27
caverac
12.8k21028
answered Nov 28 at 16:27
caverac
12.8k21028
answered Nov 28 at 16:27
caverac
12.8k21028
answered Nov 28 at 16:27
caverac
12.8k21028
12.8k21028
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
add a comment |
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
What went wrong for me was that I failed to differentiate y. Can you show, step by step, how you got $d[ln(x+c)]/dx$ to be equal to $1/(x+c)$ ?
– Ryan Cameron
Nov 28 at 16:47
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
@RyanCameron It is just the derivative of $ln$: $$frac{{rm d}ln u(x)}{{rm d} x} = frac{1}{u(x)}frac{{rm d}u}{{rm d}x}$$
– caverac
Nov 28 at 16:49
add a comment |
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Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 at 16:23