calcolare45pj

2022-03-28

Differential equations: Shouldn't the method of separation of variables miss the solutions when $x=0$?
Consider the ODE:
$x\frac{dy}{dx}=2y$ (1)
We apply the separation of variables technique to obtain:
$\frac{1}{y}dy=\frac{2}{x}dx$ (2)
which provides the general solution:
$y\left(x\right)=C{x}^{2}$
The book mentions that "the general solution (3) satisfies (1) for any value of the constant C and for all values of the variable x".
However, I believe the statement is wrong when $x=0$. By using the separation of variables, we divided by x and y to obtain (2). Thus, we make the implicit assumption that $x\ne 0$, and the separation of variables will generate solutions where $x\ne 0$.
How can y be defined for any x, when we implicitly assume that $x\ne 0$?

### Answer & Explanation

slokningol09

You are correct, as an ordinary DE the domain does not contain the vertical line $x=0$. It is a special feature of this equation that you can take two solutions $={C}_{1}{x}^{2}$ for $x<0$ and $={C}_{2}{x}^{2}$ for $x>0$ and connect them to a differentiable function at $x=0$. But there is no innate connection between the two constants, one can take any random combination.

Dixie Reed

Step 1
For a differential equation of the form $\alpha \left(x\right){y}^{\prime }\left(x\right)=f\left(x\right)\left(g\circ \left\{y\right\}\right)\left(x\right)$ in y, y need not be differentiable at any points such that $\alpha \left(x\right)=0$ in order for the equation to be satisfied everywhere else. Hence, the domain of y must initially be assumed to be $\mathrm{ℝ}\setminus {Z}_{\alpha }$, where , and it is only extended to $\mathbb{R}$ if the singularities are removable.
In your example equation, $x{y}^{\prime }\left(x\right)=2y\left(x\right)$, y is assumed implicitly to have domain $\mathbb{R}\mathrm{\setminus }\left\{0\right\}$, making the domain disconnected. The equation must be separately solved for the cases . For the case of $x<0$, there are three further cases to consider: $y<0,y=0$, which is trivial, and $y>0$. For the three cases, one has

$y\left(x\right)=0$
,
respectively. These simplify to $y\left(x\right)=-{C}_{-x}^{2}$
$y\left(x\right)=0{x}^{2}$
$y\left(x\right)={C}_{-x}^{2}$,
which are the cases such that
$y\left(x\right)={A}_{-x}^{2},,{A}_{-<}0$
$y\left(x\right)={A}_{-x}^{2},,{A}_{\equiv }0$
$y\left(x\right)={A}_{-x}^{2},,{A}_{->}0$.
Step 2
The cases can be collapsed into a single case, where ${A}_{-\in }\mathbb{R}$, and $y\left(x\right)={A}_{-x}^{2}$.
For $x>0$, the analysis is completely analogous, resulting in the most general solution to be $y\left(x\right)={A}_{+}{x}^{2}$ where ${A}_{+}\in \mathbb{R}$. Thus, the general solution family for the equation is given by
$y\left(x\right)=\left\{\begin{array}{ll}{A}_{-}{x}^{2}& x<0\\ {A}_{+}{x}^{2}& x>0\end{array}$
Of course, some solutions can be continuously and differentiably extended to $\mathbb{R}$, but this requires that ${A}_{\equiv }{A}_{+}=A$. Such solutions satisfy the differential equation in $\mathbb{R}$.

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