ODE - indeterminate coefficients

[twoflower]

Hi all,

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

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

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

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

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

$$y' = b'$$

My condition will be

$$b' = 0$$

So

$$y'' = y''' = 0$$

And I get

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

$$b = 1 - x^3$$

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

Thank you!

 

[TD]

The RHS is a polynomial of degree 3 so you're looking for a (particular) solution of the form $y = ax^3+bx^2+cx+d$. Now also find y''' and plug it into your DE, work it out and group in powers of x to identify the coefficients which will allow you to determine a,b,c,d.

 

[twoflower]

The RHS is a polynomial of degree 3 so you're looking for a (particular) solution of the form $y = ax^3+bx^2+cx+d$. Now also find y''' and plug it into your DE, work it out and group in powers of x to identify the coefficients which will allow you to determine a,b,c,d.

Thank you! I will try that. Anyway, can you see some flaw in my approach? I think it's maybe more general but should work..what do you think?

 

[TD]

Well it's not very clear to me why you would propose a lineair combination of sin and cos as a particular solution if your non-homogenous part is a polynomial. Since you take the angles 0, you end up with just a constant b but that will never be able to form the third power you have at your RHS, which is why your general solution should be of the form of a 3rd degree polynomial.

 

[twoflower]

Well it's not very clear to me why you would propose a lineair combination of sin and cos as a particular solution if your non-homogenous part is a polynomial. Since you take the angles 0, you end up with just a constant b but that will never be able to form the third power you have at your RHS, which is why your general solution should be of the form of a 3rd degree polynomial.

Well, in school, we had been told this:

If the right side is so called quasipolynom, ie. function of form

$$f(x) = e^{\alpha x}( P_1(x) \cos (\beta x) + P_2(x) \sin (\beta x))\mbox{ , }P_1\mbox{, }P_2\mbox{ polynoms}$$

Then the solution of similar form exists:

$$y(x) = x^{m}e^{\alpha x} (Q_1(x)\sin(\beta x) + Q_2(x)\cos (\beta x))$$

where $m$ is multiplicity of root $\alpha + \beta i$ of polynom $P(\lambda)$ ($m = 0$ if $m$ isn't a root)

$Q_1$ and $Q_2$ are polynoms of degree $\leq \max(\mbox{deg}(P_1), \mbox{deg}(P_2))$

I thought that expressing $y$ as $y = b$ would be ok because b can represent polynom of the desired degree. But it seems that the $a$ and $b$ in the way I use them should be really only coefficients..

 

[TD]

Oh I see, well if "b" represents a polynomial of the proper degree (here 3), then it's ok but that wasn't clear from your notation. It seemed to me that b was an unknown, just a coefficient. I think it would be better to write the complete polynomial, since as you saw: you treated b as a constant and not as a polynomial.

As far as I've been taught, the suggested polynomial Q should be of the same degree as P (which seems more logical to me actually...) not 'less or equal'.

 

[twoflower]

Oh I see, well if "b" represents a polynomial of the proper degree (here 3), then it's ok but that wasn't clear from your notation. It seemed to me that b was an unknown, just a coefficient. I think it would be better to write the complete polynomial, since as you saw: you treated b as a constant and not as a polynomial.

Great, I see it now, I really treat $b$ as if it were a constant, that's the point.

As far as I've been taught, the suggested polynomial Q should be of the same degree as P (which seems more logical to me actually...) not 'less or equal'.

Well, maybe it means that $\mbox{deg}(Q_1) = \mbox{deg}(P_1)$ and $\mbox{deg}(Q_2) = \mbox{deg}(P_2)$, that's covered by what I wrote - if you understand it in a way that "for both $P_1$ and $P_2$ holds true that their degree is less or equal to $\mbox{max(deg(}P_1\mbox{), deg(}P_2\mbox{))}$"

 

[TD]

Yes that's possible. Bottom line really is: you need to suggest a polynomial of at least the same (but exactly the same is sufficient) degree for a particular solution.

Did you manage to finish this DE?