Type:

Other

Description:

This Demonstration shows the solution to the Cauchy–Euler equation ... with initial conditions ... and ... and approximations to it using truncated series. Assume solutions have the form ... . Take the first and second derivatives of this equation and substitute back into the original equation. If the equation is to be satisfied for all ... , the coefficient of each power of ... must be zero. This gives a quadratic equation in ... with roots ... and ... . Then the coefficients ... , ... , ... , … can be determined. The two solutions for ... are: ... , ... . The final solution (plotted in blue) has the form ... , where ... and ... are determined by the initial conditions. With not too many terms it serves as a good approximation to the exact solution (plotted in red).

Subjects:

    Education Levels:

      Keywords:

      EUN,LOM,LRE4,work-cmr-id:262129,http://demonstrations.wolfram.com:http://demonstrations.wolfram.com/SeriesSolutionOfACauchyEulerEquation/,ilox,learning resource exchange,LRE metadata application profile,LRE

      Language:

      Access Privileges:

      Public - Available to anyone

      License Deed:

      Creative Commons Attribution 3.0

      Collections:

      None
      This resource has not yet been aligned.
      Curriki Rating
      'NR' - This resource has not been rated
      NR
      'NR' - This resource has not been rated

      This resource has not yet been reviewed.

      Not Rated Yet.

      Non-profit Tax ID # 203478467