We have a solution for a differential equation which is of the form:

$y(t) \LARGE = \frac{1}{C_1e^t+1}$

Since all the constant does is shift the y(t) to left or right, we can substitute $\large ce^{t} = e^{t+f(C_1)}$ and write the equation as:


How do I solve $f(C_1)$ (in other words how do I go from the first solution to the second one)?

  • 1
    $\begingroup$ Do you want to get $f(C_1)$ given $C_1$? That is just $\ln C_1$. $\endgroup$ – KittyL Dec 11 '15 at 18:10


Rewrite C1 as $\large e^{ln (C1)}$

$y(t) \LARGE = \frac{1}{e^{ln(c_1)}e^t+1}$


$y(t) \LARGE = \frac{1}{e^{t+ln(c_1)}+1}$


$f(c_1) = ln(c_1)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.