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$x\ne 0$ ?

Consider the ODE:

We apply the separation of variables technique to obtain:

which provides the general solution:

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

How can y be defined for any x, when we implicitly assume that

slokningol09

Beginner2022-03-29Added 6 answers

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

Beginner2022-03-30Added 15 answers

Step 1

For a differential equation of the form $\alpha \left(x\right){y}^{\prime}\left(x\right)=f\left(x\right)(g\circ \left\{y\right\})\left(x\right)$ 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 $\mathrm{\mathbb{R}}\setminus {Z}_{\alpha}$, where $Z}_{\alpha}{\textstyle \phantom{\rule{0.222em}{0ex}}}=\{x\in \mathbb{R}:\alpha \left(x\right)=0\$, 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 $x<0\text{}\text{and}\text{}x0$. 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

$\mathrm{ln}[-y\left(x\right)]=2\mathrm{ln}(-x)+{C}_{-}$

$y\left(x\right)=0$

$\mathrm{ln}\left[y\left(x\right)\right]=2\mathrm{ln}(-x)+{C}_{-}$,

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

$y\left(x\right)=\{\begin{array}{ll}{A}_{-}{x}^{2}& x<0\\ {A}_{+}{x}^{2}& x>0\end{array}$

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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