# Thread: [SOLVED] First order non linear DE

1. ## [SOLVED] First order non linear DE

I wonder what method to use to solve the following DE: $y'=\frac{1}{e^y -x}$ with the initial condition $y(1)=0$. As a tip they suggest to think of x as a dependent variable of y.
So I'd start with $y'(e^y -x(y))=1$. I don't think it's possible to reduce this DE into a linear one. I'm disoriented. Could you provide the name of a method to solve it? So that I check out my book/Internet and try to solve it alone.
Thanks.

2. Originally Posted by arbolis
I wonder what method to use to solve the following DE: $y'=\frac{1}{e^y -x}$ with the initial condition $y(1)=0$. As a tip they suggest to think of x as a dependent variable of y.
So I'd start with $y'(e^y -x(y))=1$. I don't think it's possible to reduce this DE into a linear one. I'm disoriented. Could you provide the name of a method to solve it? So that I check out my book/Internet and try to solve it alone.
Thanks.
They are correct try this

$\frac{dy}{dx}=\frac{1}{e^y-x}\iff \frac{dx}{dy}=e^y-x$
This gives

$x'+x=e^{y}$ this can be solved with an integrating factor $e^{y}$
Where the derivative is now with respect to y

3. I reached $y=\ln |2x-1|$ which seems to be wrong if I plug it back into the original equation.
My work: IF $=e^y$. So $xe^y=\int e^{2y}dy=\frac{e^{2y}}{2}+C$.
Thus $x=\frac{e^y}{2}+C \Rightarrow e^y=2x+C \Rightarrow y=\ln |2x+C|$ and I determined C with the initial condition $y(1)=0$.
I wonder where I went wrong.

4. Originally Posted by arbolis
I reached $y=\ln |2x-1|$ which seems to be wrong if I plug it back into the original equation.
My work: IF $=e^y$. So $xe^y=\int e^{2y}dy=\frac{e^{2y}}{2}+C$.
Thus $x=\frac{e^y}{2}+C \Rightarrow e^y=2x+C \Rightarrow y=\ln |2x+C|$ and I determined C with the initial condition $y(1)=0$.
I wonder where I went wrong.
This step is correct
$e^{y}x=\frac{1}{2}e^{2y}+C$

This is a quadratic in $e^{y}$ so set $z=e^{y}$

To solve for $xz=\frac{1}{2}z^2+C$ multiplying by 2 and cleaning up gives the equation

$z^2-2xz=\hat{C} \iff (z-x)^2=\hat{C-x^2} =$
$z=x \pm \sqrt{\hat{C}-x^2} \iff y=\ln\left(x\pm\sqrt{\hat{C}-x^2} \right)$

5. Originally Posted by TheEmptySet
This step is correct
$e^{y}x=\frac{1}{2}e^{2y}+C$

This is a quadratic in $e^{y}$ so set $z=e^{y}$

To solve for $xz=\frac{1}{2}z^2+C$ multiplying by 2 and cleaning up gives the equation

$z^2-2xz=\hat{C} \iff (z-x)^2=\hat{C-x^2} =$
$z=x \pm \sqrt{\hat{C}-x^2} \iff y=\ln\left(x\pm\sqrt{\hat{C}-x^2} \right)$
I see. I couldn't simply divide $C$ by $e^y$ and consider the result as another constant, $C$. Thanks for the help.