[EX] Solving some ODEs without any integration

[gugo82]

Exercise:

1. Let \(\Omega \subseteq \mathbb{R}^2\setminus \{\mathbf{o}\}\) be an open set made up of rays from \(\mathbf{o}=(0,0)\), let \(f,g:\Omega \to \mathbb{R}\) be two smooth \(\alpha\)-homogeneous functions with \(\alpha\neq -1\) and \((x_0,y_0)\in \Omega\) s.t. \(g(x_0,y_0)\neq 0\).
Prove that if:
\[
\tag{1}
f_y (x,y) = g_x (x,y)
\]
in \(\Omega\), then any solution of the Cauchy's problem:
\[
\tag{P}
\begin{cases}
g(x,y(x))\ y^\prime (x) = - f(x,y(x))\\
y(x_0) = y_0
\end{cases}
\]
is implicitly defined by the following equation:
\[
\tag{2}
x\ f(x,y) + y\ g(x,y) = x_0\ f(x_0,y_0) + y_0\ g(x_0 ,y_0)
\]
in a suitable neighbourhood of the initial data \((x_0,y_0)\).

2. Solve:
\[
\tag{3}
\begin{cases}
2x\ y(x)\ y^\prime (x) = x^2 -y^2(x)\\
y(\sqrt{3}) = 1\; .
\end{cases}
\]

[gugo82]

"gugo82":1. Let \( \Omega \subseteq \mathbb{R}^2\setminus \{\mathbf{o}\} \) be an open set made up of rays from \( \mathbf{o}=(0,0) \), let \( f,g:\Omega \to \mathbb{R} \) be two smooth \( \alpha \)-homogeneous functions with \( \alpha\neq -1 \) and \( (x_0,y_0)\in \Omega \) s.t. \( g(x_0,y_0)\neq 0 \).
Prove that if:
\[ \tag{1} f_y (x,y) = g_x (x,y) \]
in \( \Omega \), then any solution of the Cauchy's problem:
\[ \tag{P} \begin{cases} g(x,y(x))\ y^\prime (x) = - f(x,y(x))\\ y(x_0) = y_0 \end{cases} \]
is implicitly defined by the following equation:
\[ \tag{2} x\ f(x,y) + y\ g(x,y) = x_0\ f(x_0,y_0) + y_0\ g(x_0 ,y_0) \]
in a suitable neighbourhood of the initial data \( (x_0,y_0) \).

Let:
\[
\Phi (x,y) := x\ f(x,y) + y\ g(x,y)\; .
\]
Since the functions \((x,y)\mapsto x\) and \((x,y)\mapsto y\) are \(1\)-homogeneous, both products \(x\; f(x,y)\) and \(y\; g(x,y)\) are \(\alpha +1\)-homogeneous functions; hence, \(\Phi\) is an \(\alpha +1\)-homogenous function which is differentiable in \(\Omega\) whose derivatives satisfy:
\[
\begin{split}
\Phi_x (x,y) &= f(x,y) + x\ f_x(x,y) + y\ g_x(x,y)\\
&= f(x,y) + x\ f_x(x,y) + y\ f_y(x,y)\\
&= f(x,y) + \alpha\ f(x,y)\\
&= (\alpha +1)\ f(x,y)\\
\Phi_y(x,y) &= g(x,y) + x\ f_y(x,y) + y\ g_y(x,y))\\
&= g(x,y) + x\ g_x(x,y) + y\ g_y(x,y)\\
&= g(x,y) + \alpha\ g(x,y)\\
&= (\alpha +1)\ g(x,y)
\end{split}
\]
with \(\alpha +1\neq 0\), because of Euler's Theorem on positively homogeneous functions. Therefore the ODE in (P) rewrites:
\[\tag{I}
\Phi_x(x,y(x))+ \Phi_y(x,y(x))\ y^\prime (x) =0\; .
\]

Now, assume that (P) has at least a solution (which is indeed the case, because Peano's Existence Theorem applies).
Upon setting:
\[
\phi (x) := \Phi (x,y(x))\; ,
\]
from (I) we infer:
\[
\phi^\prime (x) =0
\]
and finally:
\[
\tag{II}
\phi (x) = \phi (x_0)
\]
everywhere in the interval where the maximal solution of (P) is defined. But (II) is in fact equivalent to (2), so we have the claim.

[gugo82]

Nice try... That could be a way to solve problem (3) using the classical Frobenius method (a.k.a. method of power series).
There are some passages missing, but nothing you cannot fill by yourself.

Another way consists in using formula (2) from part 1.

In fact, the ODE in (3) is of the type (P) with coefficients \(f(x,y) := y^2-x^2\) and \(g(x,y) := 2xy\) satisfying (1) everywhere, therefore, any solution of the Cauchy's problem (3) satisfies:
\[
\begin{split}
x\ (y^2 - x^2) + y\ (2xy) = \sqrt{3}\ (1-3) + 1\ (2\sqrt{3})\qquad &\Leftrightarrow \qquad x\ (3y^2-x^2) =0\\
&\Leftrightarrow \qquad x\ (\sqrt{3}\ y - x)\ (\sqrt{3}\ y + x) =0\; ;
\end{split}
\]
the latter equation forces \(y=\frac{1}{\sqrt{3}}\ x\) around \((x_0,y_0)=(\sqrt{3} ,1)\), hence the function:
\[
y(x;\sqrt{3} ,1) = \frac{1}{\sqrt{3}}\ x\quad ,\ x\in ]0,\infty[
\]
is the maximal solution of (3).[nota]The domain of the solution is \(]0,\infty[\) because the ODE in (3) degenerates when \(x=0\) or \(y=0\); therefore, the (maximal) domain \(]X_-,X^+[\) of \(y(\cdot ;\sqrt{3} ,1)\) must have \(X_->0\) and the solution itself had to be strictly positive in \(]X_-,X^+[\).[/nota]

Any idea for part 1?

Hint: Use Euler's theorem for homogeneous functions.

[Sk_Anonymous]

2) A partial solution.
A) $xyy'=1/2x^2-1/2y^2$
From (A) :
$y'(x)={x^2-y^2}/{2xy}->y'(sqrt 3)={3-1}/{2sqrt3}=1/{sqrt3}$
Deriving (A) :
B) $yy'+x(y')^2+xyy''=x-yy'$
From (B):
$y''(x)={x-2yy'-x(y')^2}/{xy}-> y''(sqrt 3)={sqrt 3-{2}/{sqrt3 }-{sqrt3}/{3}}/{sqrt3 }=0$
Deriving (B) :
C) $(y')^2+yy''+(y')^2+2xy'y''+yy''+xy'y''+xyy'''=1-(y')^2-yy''$
From (C) :
$y'''(x)={1-3(y')^2-3yy''-3xyy''}/{xy}->y'''(sqrt3)={1-3/3-0-0}/{sqrt3}=0$
and so on.
Generalizing (?), result
$y(sqrt3)=1,y'(sqrt 3)=1/{sqrt3}, y^{(i)}(sqrt 3)=0, i>=2$
Then : ${y(x)=x/{\sqrt 3} } $