MHB Finding the Solution to a Differential Equation with Reciprocal Relationship

[karush]

$\textsf{Find the solution of:}$
$\displaystyle\frac{dy}{dx}=\frac{1}{e^y-x}$
ok i kinda don't know what the first step is
was going to multiply both sides by the denominator but

 

[MarkFL]

I would begin here with the substitution:

$$u=e^y\implies u'=uy'$$

And so the ODE becomes:

$$\frac{1}{u}u'=\frac{1}{u-x}$$

Or:

$$x'+\frac{1}{u}x=1$$

Now you have a linear equation in $x$...can you proceed?

 

[MarkFL]

We could also simply state:

$$\d{x}{y}=e^y-x$$

$$\d{x}{y}+x=e^y$$

Now, our integrating factor is:

$$\mu(y)=\exp\left(\int\,dy\right)=e^y$$

And we then obtain:

$$\frac{d}{dy}\left(e^yx\right)=e^{2y}$$

Integrate w.r.t $y$:

$$e^yx=\frac{1}{2}e^{2y}+c_1$$

Now, we have a quadratic in $e^y$ which cay be written in standard form as:

$$e^{2y}-2xe^{y}-c_2=0$$ ($c_2=-2c_1$)

Applying the quadratic formula, we get:

$$e^y=\frac{2x\pm\sqrt{4x^2+4c_2}}{2}=x\pm\sqrt{x^2+c_2}$$

And this implies:

$$y(x)=\ln\left|x\pm\sqrt{x^2+c_2}\right|$$

 

[karush]

We could also simply state:
$$\d{x}{y}=e^y-x$$
$$\d{x}{y}+x=e^y$$
Now, our integrating factor is:
$$\mu(y)=\exp\left(\int\,dy\right)=e^y$$
And we then obtain:
$$\frac{d}{dy}\left(e^yx\right)=e^{2y}$$
Integrate w.r.t $y$:
$$e^yx=\frac{1}{2}e^{2y}+c_1$$

$\textsf{ok the book answer was}$
$$x=ce^{-y}+\frac{1}{2}e^y$$
$\textsf{which would be derived from}$
$$e^yx=\frac{1}{2}e^{2y}+c_1$$
$\textsf{by dividing thru by $e^y$}$

 

[MarkFL]

$\textsf{ok the book answer was}$
$$x=ce^{-y}+\frac{1}{2}e^y$$
$\textsf{which would be derived from}$
$$e^yx=\frac{1}{2}e^{2y}+c_1$$
$\textsf{by dividing thru by $e^y$}$

That's fine giving $x$ as a function of $y$...I just chose to give $y$ as a function of $x$ since the original equation has $x$ as the independent variable. :)