Suppose I am given a Cauchy-Euler form second order differential equation x^2(d^2y/dx^2)+x(dy/dx)+y=f(x).

Lillie Pittman 2022-07-16 Answered
Suppose I am given a Cauchy-Euler form second order differential equation
x 2 d 2 y d x 2 + x d y d x + y = f ( x ) .
The usual textbook method for solving the Cauchy Euler equation is to blackuce it to a linear differential equation with constant coefficients by the transformation x = e t . But I have a fundamental doubt here, we know that e t > 0 t R . But when we are using the above transformation we are subconsciously assuming x > 0. How does this make sense?

Substituting t = ln ( x ) also makes no difference as ln is defined on R > 0 . So I now doubt the validity of the method followed to solve the Cauchy-Euler equation.

Can someone give me a proper explanation of what exactly going on here and why the process is valid?
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Answers (2)

Ali Harper
Answered 2022-07-17 Author has 16 answers
The Euler-Cauchy equation has a singularity at x = 0. This singularity splits the domain of the ODE, which in turn limits any solution of the ODE to a sub-interval of either x > 0 or x < 0, depending on the initial condition. In the unusual case where the initial condition is at some negative x 0 , you can of course also use the substitution x = e t or more generally x = x 0 e t .
So with u ( t ) = y ( x 0 e t ) one gets u ( t ) = y ( x 0 e t ) x 0 e t = x y ( x ) and u ( t ) = y ( x 0 e t ) ( x 0 e t ) 2 + y ( x 0 e t ) x 0 e t = x 2 y ( x ) + x y ( x ). Here one can see that the effect of the substitution is rather independent of x 0 and its sign.
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Emmanuel Pace
Answered 2022-07-18 Author has 6 answers
The Eular equation x 2 y + x y + y = 0         ( 1 ) can be solved y assuming y = x m (Euler's substitution: ES), then we get m = ± i, so the solution is y = C 1 x i + C 2 x i as
x i = e ln x i = e i ln x = cos ( ln x ) + i sin ( ln x ) .
So the solution of (1) can also be qrutten as y = D 1 sin ln x + D 2 cos ln x. So so far no problem with ES.

Alternatively one can claim that y 1 = sin ln x , y 2 = cos ln x are two linearly independent solutions of (1). However the solution of the in-homogeneous ODE:
x 2 Y + x Y + Y = f ( x )         ( 2 )
can be obtained by the method of variation of parameters using y 1 , y 2 .
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