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
$\endgroup$ 12 Answers
$\begingroup$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:$$\frac{\mbox{d}}{\mbox{d}x}\ln(x+c)=\frac{1}{x+c}\underbrace{\frac{\mbox{d}}{\mbox{d}x}\left(x+c\right)}_{=1}=\frac{1}{x+c}$$
$\endgroup$ 3 $\begingroup$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
$\endgroup$ 2
