Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

I can't figure out where I've doing mistake in this quite simple problem:

[tex]

y''' - y = x^3 - 1

[/tex]

FSS = [itex][e^{x}, e^{-\frac{1}{2}x} \cos \left(\frac{\sqrt{3}}{2}x\right), e^{-\frac{1}{2}x} \sin \left(\frac{\sqrt{3}}{2}x\right)][/itex]

Then, because on the right side there is quasipolynom, I can solve it using the method of indeterminate coefficients.

So I look for solution of form

[tex]

y(x) = a\sin (0x) + b\cos (0x) = b

[/tex]

[tex]

y' = b'

[/tex]

My condition will be

[tex]

b' = 0

[/tex]

So

[tex]

y'' = y''' = 0

[/tex]

And I get

[tex]

y''' - y = -b = x^3 - 1

[/tex]

[tex]

b = 1 - x^3

[/tex]

Anyway, this is not correct...can you see where am I doing the mistake?

Thank you!

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# ODE - indeterminate coefficients

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