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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: ODE - indeterminate coefficients

**Physics Forums | Science Articles, Homework Help, Discussion**