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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 ?

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There is no difference. You use different constants. Put $D=-\ln C_1$ in your solution. –  Artem Jan 28 '13 at 16:50

2 Answers 2

up vote 5 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.

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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 $c_2 = -Ae^D$.

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$c_2=-Ae^D{}{}$. –  Cameron Buie Jan 28 '13 at 19:21

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