Consider the differential equation $y'=f(x,y)$ where $f$ is homogeneous of degree 0.
Thus, $f(tx,ty)=f(x,y)$ for all $x,y$ and all $t \neq 0$.
Let $t=\frac{1}{x}$. Then
$$y'=f\left (1,\frac{y}{x}\right)\tag{8.54}$$
The appearance of the quotient $y/x$ on the right suggests that we introduce a new unknown function $v$ where $v=y/x$. Then $y=vx$, $y'=v'x+v$, and this substitution transforms (8.54) into
$$v'x+v=f(1,v)$$
$$x\frac{dv}{dx}=f(1,v)-v$$
This last equation is a first-order separable equation for $v$.
When we say $v=y/x$ isn't $v$ a function of both $y$ and $x$?
When we differentiate $y=vx$, why don't we use a partial derivative $\partial v/\partial x$?
We can rewrite as
$$\frac{1}{f(1,v)-v}v'=\frac{1}{x}$$
Let $R(v)=f(1,v)-v$, $A(v)=\frac{1}{R(v)}$, and $Q(x)=\frac{1}{x}$.
Then the differential equation is
$$A(v(x))v'(x)=Q(x)\tag{1}$$
Consider the following theorem
Theorem 8.10 Let $y=Y(x)$ be any solution of the separable differential equation
$$A(y)y'=Q(x)\tag{8.45}$$
such that $Y'$ is continuous on an open interval $I$. Assume that both $Q$ and the composite function $A\circ Y$ are continuous on $I$. Let $G$ be any primitive of $A$, that is, any function such that $G'=A$. Then the solution $Y$ satisfies the implicit formula
$$G(y)=\int Q(x)dx + C\tag{8.46}$$
for some constant $C$ Conversely, if $y$ satisfies (8.46) then $y$ is a solution of $(8.45)$.
I'd like to apply this theorem to solve (1).
$Q(x)$ is continuous on $\mathbb{R}\backslash\{0\}$.
$A(v)$ is continuous on $\mathbb{R}\backslash\{v: v\neq f(1,v)\}$
We assume a solution $v(x)$ is such that $v'(x)$ is continuous on an open interval $I$ we're interested in.
What is $I$ after all?
If we integrate (1) we end up with
$$\frac{1}{2}\ln{(1+v^2)}+\arctan{(v)}=\ln{|x|}+C$$
and if we sub in $v=\frac{y}{x}$ we get
$$\frac{1}{2}\ln{(x^2+y^2)}+\arctan{\left (\frac{y}{x}\right )}=C$$
which is an implicit formula for the solution $y(x)$ of (8.54).
Here is a specific example
$$(x+1)y'+y^2=0$$
$$\frac{1}{y^2}y'=-\frac{1}{x+1}$$
$$A(y(x))y'(x)=Q(x)$$
$Q(x)$ is continuous on $I_1=\mathbb{R}\backslash\{0\}$.
$A(y(x))$ is continuous on $I_2=\mathbb{R}\backslash \{x:y(x)\neq 0\}$.
If we assume we are working on the interval $I_1 \cap I_2$ then we can integrate both sides to obtain
$$-y^{-1}=-\ln{|x+1|}+C$$
$$y(x)=\frac{1}{\ln{|x+1|}+C}$$
which is continuous everywhere except when $\ln{|x+1|}=-C$.
Every such solution is always $\neq 0$.
Thus, going back to our statement about the continuity of $A(y(x))$ we have that $A(y(x))$ is continuous on all of $\mathbb{R}$.
Thus, it seems that the relevant open interval on which we have solutions is $I=I_1\cap I_2=\mathbb{R}\backslash(\{0\}\cup\{x: \ln{|x+1|}=-C\})$.
Here is my attempt at answering my own question
We find $I_1$ where $Q(x)$ is continuous
We find $I_2$ where $A(y(x)$ is continuous. We don't know what $I_2$ is exactly at this point.
We integrate the separable DE on $I_1\cup I_2$ and obtain a family of solutions.
We can now know exactly what $I_2$ is.
We also can see where the solution is not continuous, and hence where it is continuous, call this interval $I_3$.
Thus, we know what $I_1\cap I_2 \cap I_3$ is and hence where our solution works.
Is this correct?
- The solutions