Solving Euler-Cauchy Equation: General Solution with Euler Method

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The discussion focuses on solving the Euler-Cauchy equation represented as (10x^2D^2 - 20xD + 22.4I)y = 0. The general solution can be approached by assuming a solution of the form y = x^r, leading to the characteristic equation 10r^2 + 10r + 22.4 = 0, which yields complex solutions. An alternative method involves changing the variable to x = ln(t), transforming the equation into one with constant coefficients. Additionally, the Euler method can be adapted for numerical solutions by converting the second-order equation into a system of first-order equations.

PREREQUISITES
  • Understanding of Euler-Cauchy equations
  • Familiarity with characteristic equations
  • Knowledge of complex numbers and their interpretation in solutions
  • Basic principles of numerical methods, specifically Euler's method
NEXT STEPS
  • Study the derivation of characteristic equations for second-order linear differential equations
  • Learn about the transformation of variables in differential equations, particularly x = ln(t)
  • Explore numerical methods for solving higher-order differential equations, focusing on the Euler method
  • Investigate the interpretation of complex solutions in the context of differential equations
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Mathematicians, engineering students, and anyone involved in solving differential equations, particularly those interested in Euler-Cauchy equations and numerical methods.

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how to find the general solution for the equation
(10x^2D^2-20xD+22.4I)y=0...no idea to find it by using euler-method...help..sos
 
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Unfortunately, Euler's name is attached to to many things!

That is an "Euler" or "Euler-type" equation. One way of solving it is to try a solution of the form [itex]y= x^r[/itex] for some number r. Then [itex]y'= rx^{r-1}[/itex] and [itex]y''= r(r-1)x^{r-2}[/itex]. Putting those into the equation gives
[itex](10r(r-1)+ 20r+ 22.4)x^r= 0[/itex] and, since [itex]x^r[/itex] is not always 0,
[itex]10r(r- 1)+ 20r+ 22.4= 10r^2+ 10r+ 22.4= 0[/itex]

That will have complex solutions so you would have to interpret [itex]x^{a+ bi}[/itex] in terms of sine and cosine.

Another way of doing that would be to make the change of variable x= ln(t) which converts the "Euler-type" equation to an equation with constant coefficents having the same characteristic equation.

Or, since you say "Euler method" are you referring Euler's numerical method for solving equations? Strictly speaking that applies to first order equations but you can convert this second order equation to a pair of first order equations:

Let y= Dy so that [itex]D^2 y= Du[/itex] any your equation becomes
[itex]x^2Du- 20xDy+ 22.4y= 0[/itex]
and [itex]Dy= u[/itex]
You can set up an "double" Euler-method solver that, given the values of u and y at a each step, solves Dy= u for y and then puts both of those values into the other equation to solve for y.
 
but the answer given to the equation is C1x^1.4 + C2X^1.6,its no complex solutions...and i just learn the euler cauchy equation only..
 

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