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(10x^2D^2-20xD+22.4I)y=0...no idea to find it by using euler-method...help..sos

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- Thread starter xw0927
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- #1

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(10x^2D^2-20xD+22.4I)y=0...no idea to find it by using euler-method...help..sos

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HallsofIvy

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That

[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

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.

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