Initial value problem Euler equation

lisa92
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Question:

Find y as a function of x:

x^2 y'' + 8 x y' - 18 y = x^8

y(1)=3, y'(1)=2


Attempted solution:

I found the general equation to be Ax^(-9)+Bx^2+Cx^8.
However when I try to solve the initial value problem for this equation I have 3 unknowns.
 
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Welcome to PF, lisa92! :smile:

Did you substitute your solution in the DE?
If you do, you'll find you have a 3rd equation.
 
write
(x2D2+8 x D-18)=(xD+9)(xD-2)
or change variables u=x2 y

your general solution is extraneous check it in the equation to eliminate
 


That's an "Euler type" equationl. The change of variable x= ln(t) changes it to a differential equation with constant coefficients which may be simpler.

If you do not wish to do it that way, what did you get as a solution to the associated homogeneous equation?
 


Check out Euler-Cauchy equations at the bottom of this link:

http://sosmath.com/tables/diffeq/diffeq.html

This will help you obtain the solution to the homogeneous equation.

The particular solution will probably be y = k * x^8 + y (homogeneous)
 
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