Non-linear second order ODE

Question

I have to solve $$ y''(x)+(y'(x))^2=y'(x). $$

Using $ y'(x)=z $, I can write $$\int \frac{1}{z-z^2}dz=\int dx $$

So: $$\frac{1}{z(1-z)}=\frac{A}{z}+\frac{B}{1-z}$$ leads to

$$ \int \frac{1}{z(1-z)}dz=\int \frac{1}{z}dz+\int \frac{1}{1-z}dz= \ln(z)-\ln(1-z)$$ $$\Rightarrow \ln\left(\frac{z}{1-z}\right)=x+c $$ $$\Rightarrow z=\frac{e^{x+c}}{1+e^{x+c}}=y'$$ $$\Rightarrow y=\int \frac{e^{x+c}}{1+e^{x+c}}dx $$

Now, calling $e^{x+c}=t$:

$$y=\int\frac{t}{1+t}\cdot\frac{dt}{t}=\ln(1+t)\Rightarrow \ln(1+e^{x+c})$$

I checked the calculations and i thought was right but WolframAlpha says that the result is $\ln(c_1+e^x)+c_2$. What am I doing wrong?

Thanks for any help!

Answer 1

You didn't do anything wrong. The answer you got is $\ln(1+e^{x+c_1})+c_2$. It is equal to:

\begin{align} \ln(1+e^{x+c_1})+c_2 &= \ln(e^{c_1}(e^x+\frac{1}{e^{c_1}}))+c_2 \\ &= \ln(e^{c_1})+\ln(e^x+\frac{1}{e^{c_1}})+c_2 \\ &= \ln(\frac{1}{e^{c_1}}+e^x)+(c_1+c_2). \end{align}

Now let $d_1=\frac{1}{e^{c_1}}$, $d_2=c_1+c_2$ and you will get the answer from wolfram.

Answer 2

Apart from missing the absolute sign for log. You have the following issues $$ y = \int \frac{1}{1+t}dt \to \ln(1+t) + C_2 $$ then couple this with the following. $$ \log(1 + \mathrm{e}^{x+c}) = \log(1 + A\mathrm{e}^{x}) = \log A\left(\frac{1}{A} + \mathrm{e}^{x}\right) = \log A + \log \left(\frac{1}{A} + \mathrm{e}^{x}\right) $$

Answer 3

Hint.

Calling $z = y'$ we have

$$ z' + z^2-z = 0 $$

Making now $u = \frac 1z\Rightarrow u'+u-1=0\Rightarrow u = 1+C_0 e^{-x}$ then

$$ z = \frac{1}{u} = \frac{e^x}{e^e+C_0}\Rightarrow y = \int z dx = \ln(e^x+C_0) + C_1 $$

Answer 4

Note that $$ y=\int \frac{e^{x+c}}{1+e^{x+c}}dx$$

$$t=e^{x+c} \implies dt =e^{x+c}dx$$

$$ y=\int \frac{e^{x+c}}{1+e^{x+c}}dx =\int \frac{dt}{1+t} $$

$$ = \ln(1+t)+c_2 = \ln(1+e^{x+c} ) +c_2$$

Their answer $$ ln(c_1+e^x)+c_2 = \ln c_1(1+e^x/{c_1})+c_2=\ln c_1 + ln (1+c_1^{-1} e^x)+c_2$$ is equivalent to your solution via renaming the constants.