$dy/dx=e^{-y}$ differential equation












0














I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:



enter image description here



However that is not a correct solution. How can I derrive the true solution?



This is not correct as:



enter image description here



I forgot to write ln, the right side hasn't changed



I used TI-Nspire CX CAS student software to come to that conclusion:
enter image description here



Weird it's given me a wrong answer






differential-equations







share|cite|improve this question
























  • Why do you believe this is not correct? Mind the notation in the one but last line, though.
    – StackTD
    Nov 28 '18 at 16:23


















0














I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:



enter image description here



However that is not a correct solution. How can I derrive the true solution?



This is not correct as:



enter image description here



I forgot to write ln, the right side hasn't changed



I used TI-Nspire CX CAS student software to come to that conclusion:
enter image description here



Weird it's given me a wrong answer






differential-equations







share|cite|improve this question
























  • Why do you believe this is not correct? Mind the notation in the one but last line, though.
    – StackTD
    Nov 28 '18 at 16:23
















0












0








0







I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:



enter image description here



However that is not a correct solution. How can I derrive the true solution?



This is not correct as:



enter image description here



I forgot to write ln, the right side hasn't changed



I used TI-Nspire CX CAS student software to come to that conclusion:
enter image description here



Weird it's given me a wrong answer






differential-equations







share|cite|improve this question















I am trying to solve the following differential-equation: $$dy/dx=e^{-y}$$
This is what I tried:



enter image description here



However that is not a correct solution. How can I derrive the true solution?



This is not correct as:



enter image description here



I forgot to write ln, the right side hasn't changed



I used TI-Nspire CX CAS student software to come to that conclusion:
enter image description here



Weird it's given me a wrong answer






differential-equations




differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 16:48

























asked Nov 28 '18 at 16:21









Ryan Cameron

397




397












  • Why do you believe this is not correct? Mind the notation in the one but last line, though.
    – StackTD
    Nov 28 '18 at 16:23




















  • Why do you believe this is not correct? Mind the notation in the one but last line, though.
    – StackTD
    Nov 28 '18 at 16:23


















Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 '18 at 16:23






Why do you believe this is not correct? Mind the notation in the one but last line, though.
– StackTD
Nov 28 '18 at 16:23












2 Answers
2






active

oldest

votes


















2















This is not correct as:



enter image description here




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:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$






share|cite|improve this answer























  • 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 '18 at 16:47










  • @RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
    – Andrew Li
    Nov 28 '18 at 16:49












  • @RyanCameron I added a bit at the end of my answer.
    – StackTD
    Nov 28 '18 at 16:50



















2














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






share|cite|improve this answer





















  • 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 '18 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 '18 at 16:49













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2















This is not correct as:



enter image description here




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:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$






share|cite|improve this answer























  • 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 '18 at 16:47










  • @RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
    – Andrew Li
    Nov 28 '18 at 16:49












  • @RyanCameron I added a bit at the end of my answer.
    – StackTD
    Nov 28 '18 at 16:50
















2















This is not correct as:



enter image description here




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:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$






share|cite|improve this answer























  • 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 '18 at 16:47










  • @RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
    – Andrew Li
    Nov 28 '18 at 16:49












  • @RyanCameron I added a bit at the end of my answer.
    – StackTD
    Nov 28 '18 at 16:50














2












2








2







This is not correct as:



enter image description here




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:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$






share|cite|improve this answer















This is not correct as:



enter image description here




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:
$$frac{mbox{d}}{mbox{d}x}ln(x+c)=frac{1}{x+c}underbrace{frac{mbox{d}}{mbox{d}x}left(x+cright)}_{=1}=frac{1}{x+c}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 '18 at 16:50

























answered Nov 28 '18 at 16:28









StackTD

22.3k2049




22.3k2049












  • 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 '18 at 16:47










  • @RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
    – Andrew Li
    Nov 28 '18 at 16:49












  • @RyanCameron I added a bit at the end of my answer.
    – StackTD
    Nov 28 '18 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 '18 at 16:47










  • @RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
    – Andrew Li
    Nov 28 '18 at 16:49












  • @RyanCameron I added a bit at the end of my answer.
    – StackTD
    Nov 28 '18 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 '18 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 '18 at 16:47












@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 '18 at 16:49






@RyanCameron Derivative of $ln |u|$ is $1/u , mathrm du$...
– Andrew Li
Nov 28 '18 at 16:49














@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 '18 at 16:50




@RyanCameron I added a bit at the end of my answer.
– StackTD
Nov 28 '18 at 16:50











2














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






share|cite|improve this answer





















  • 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 '18 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 '18 at 16:49


















2














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






share|cite|improve this answer





















  • 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 '18 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 '18 at 16:49
















2












2








2






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






share|cite|improve this answer












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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 28 '18 at 16:27









caverac

13.8k21030




13.8k21030












  • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 16:49




















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