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 &gt; 0 ∀ t ∈ R . But when we are using the above transformation we are subconsciously assuming x &gt; 0. How does this make sense?Substituting t = ln ⁡ ( x ) also makes no difference as ln is defined on R &gt; 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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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    &amp;gt;  0   ∀  t  ∈      R  . But when we are using the above transformation we are subconsciously assuming   x  &amp;gt;  0. How does this make sense?Substituting   t  =  ln  ⁡  (  x  ) also makes no difference as ln is defined on             R              &amp;gt;      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?

Ali Harper · Accepted Answer

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  &amp;gt;  0 or   x  &amp;lt;  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.

Emmanuel Pace · Accepted Answer

The Eular equation       x    2        y    ″    +  x      y    ′    +  y  =  0              (  1  ) can be solved y assuming   y  =      x    m   (Euler&#039;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  .

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

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