Differential equations: Shouldn't the method of separation of



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Differential equations: Shouldn't the method of separation of variables miss the solutions when x=0?
Consider the ODE:
xdydx=2y (1)
We apply the separation of variables technique to obtain:
1ydy=2xdx (2)
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 x=0. By using the separation of variables, we divided by x and y to obtain (2). Thus, we make the implicit assumption that x0, and the separation of variables will generate solutions where x0.
How can y be defined for any x, when we implicitly assume that x0?

Answer & Explanation



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 =C1x2 for x<0 and =C2x2 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

Dixie Reed

Beginner2022-03-30Added 15 answers

Step 1
For a differential equation of the form α(x)y(x)=f(x)(g{y})(x) in y, y need not be differentiable at any points such that α(x)=0 in order for the equation to be satisfied everywhere else. Hence, the domain of y must initially be assumed to be Zα, where Zα ={xR:α(x)=0}, and it is only extended to R if the singularities are removable.
In your example equation, xy(x)=2y(x), y is assumed implicitly to have domain R{0}, making the domain disconnected. The equation must be separately solved for the cases x<0 and x>0. 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
ln[y(x)]=2ln(x)+C ,
respectively. These simplify to y(x)=Cx2
which are the cases such that
Step 2
The cases can be collapsed into a single case, where AR, and y(x)=Ax2.
For x>0, the analysis is completely analogous, resulting in the most general solution to be y(x)=A+x2 where A+R. Thus, the general solution family for the equation is given by
Of course, some solutions can be continuously and differentiably extended to R, but this requires that AA+=A. Such solutions satisfy the differential equation in R.

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