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:

\(\displaystyle u=e^y\implies u'=uy'\)

And so the ODE becomes:

\(\displaystyle \frac{1}{u}u'=\frac{1}{u-x}\)

Or:

\(\displaystyle x'+\frac{1}{u}x=1\)

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

 

[MarkFL]

We could also simply state:

\(\displaystyle \d{x}{y}=e^y-x\)

\(\displaystyle \d{x}{y}+x=e^y\)

Now, our integrating factor is:

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

And we then obtain:

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

Integrate w.r.t $y$:

\(\displaystyle e^yx=\frac{1}{2}e^{2y}+c_1\)

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

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

Applying the quadratic formula, we get:

\(\displaystyle e^y=\frac{2x\pm\sqrt{4x^2+4c_2}}{2}=x\pm\sqrt{x^2+c_2}\)

And this implies:

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

 

[karush]

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

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

 

[MarkFL]

$\textsf{ok the book answer was}$
$$x=ce^{-y}+\frac{1}{2}e^y$$
$\textsf{which would be derived from}$
\(\displaystyle e^yx=\frac{1}{2}e^{2y}+c_1\)
$\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. :)