Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Solve $y^{\prime \prime}-(y^{\prime})^2-y^{\prime}=0$. I use $$u=\frac{dy}{dx}$$ to transform the DE into $$\frac{du}{dx}-u^2-u=0$$. I know that this is an Bernoulli equation with $n=2$. I get the final solution is $$y=-ln|1-Ae^x|+D$$ where $A=+-e^c$. But my lecturer's answer is $$y=-ln|C_1+c_2e^x|$$. May I know what is the difference my answer and my lecturer answer ?

share|cite|improve this question
There is no difference. You use different constants. Put $D=-\ln C_1$ in your solution. – Artem Jan 28 '13 at 16:50
up vote 6 down vote accepted

The answers are the same. $D$ is an arbitrary constant so let $D=\ln(D')$ for arbitrary $D'$ and $e^c$ is just another constant so $e^c=c'$ for an arbitrary $c'$, and so now $$y=-ln|1-Ae^x|+D=-ln|1-c'e^x|+\ln(D')=-\ln(D' -D' c' e^x|$$ Let $c_1=D'$ and $c_2=c'D'$ and you're done.

share|cite|improve this answer

You can see that the two answers are equivalent as follows: \begin{align} -\ln |1 - Ae^x| + D & = -\ln |1-Ae^x| + \ln e^D \\ & = -\ln e^D|1 - Ae^x| \\ & = - \ln |e^D - e^D Ae^x| \\ & = -\ln | c_1 + c_2 e^x|,\end{align} where $c_1 = e^D$ and $c_2 = -Ae^D$.

share|cite|improve this answer
$c_2=-Ae^D{}{}$. – Cameron Buie Jan 28 '13 at 19:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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