I have:
$$xy'= x^2+y^2$$
I tried to separate variables but it did not work, checked if it could be homogeneous equation but it is not (obiviously), all my transformations did not give me linear equation, so I have no clue how to approach it now :(
$\begingroup$
$\endgroup$
1
-
$\begingroup$ According to Mathematica, the simplest solution is in terms of Bessel functions, I'm afraid. $\endgroup$– Patrick StevensOct 31, 2017 at 18:50
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
It is a Riccati equation.
Let $v=\frac y x$. Then $v$ satisfies the equation $$v' = v^2 -\frac v x +1$$
Substituting $v = \frac{-u'}{u}$, we have that $u$ satisfies: $$ u'' + \frac{u'}{x} +u=0 $$
By solving this, which is a Bessel equation though, we will then have that $y = \frac{-u'}{ux}$ satisfies the initial equation.
The Bessel equation $x^2u'' +xu'+x^2u=0$, has as solutions the Bessel functions.
