Homogenous solution of a differential equation

[arhzz]

Homework Statement

Solve the differential equation

Relevant Equations

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Hello ! I need to solve this diffrential equation.

$$ y^{(4)} + 2y'' + y = 0 $$

First I wanted to find the homogenous solution,so I built the characteristic polynomial ( not sure if u say it so in english as well).I did that like this

$$\lambda^4 +2\lambda^2+1 = 0 $$. The solutins should be $$ \lambda_1 = i , \lambda_2 = -i $$ Now I used the approach to convert these complex solutions into sin and cos and I got this to be my homogenous solution.

$$y_h = c_1cos(x) + c_2sin(x) $$ Now this is not right,accorindg to wolfram alpha the solution should be

$$ y_h = (c_1+c_2x)cosx + (c_3+c_4x)sinx $$ Now the first thing that came to mind is multiplicty (not sure if it is the right word in english) but basically when the same solution appears more than once.But than I found this differential equation that I solved earlier today

$$ y'' +2y = 0 $$ Here the solution was +2i and -2i,and yet the homogenous solution had only 2 constants,
$$ y_h = c_1cos(2x)+c_2sin(2x) $$ and according to wolframalpha this is correct.I am obviously missing something but I am not sure what.

Unrelated note: The "preview" button is not working lately for me and it is incredibly annoying,whenever I click on it it does not show me a preview so I am writing "blindly" into LaTeX.Also when I reload the page something it buggs out not letting me write further,and latex literally not wanting to delete itself leading to very akward looking posts.Has any script been update,that I need to update on my browser? Because this started happening recently and its making life really annoying.

Thanks!

 

[Office_Shredder]

Multiplicity is multiplicity of the roots. If for example the characteristic polynomial is $(\lambda -i)^2(\lambda+i)$ then $i$ is a root of the polynomial of degree two since that's the power on $(\lambda-i)$, and $-i$ would have multiplicity 1.

In your first polynomial the roots have multiplicity two, in your second example they have multiplicity 1 (try factoring it)

 

[anuttarasammyak]

The solution of
$$(D^2+1) y =0,$$
$$y=c_1 e^{ix}+c_2 e^{-ix}$$
of course satisfies the given equation
$$(D^2+1)^2 y =0$$

So y satisfying
$$(D^2+1) y = c_1 e^{ix}+c_2 e^{-ix}$$
is also a solution of the given equation. Say y= $(c_1 e^{ix}+c_2 e^{-ix})z$,
$$(D^2+1) y = 2z'( ic_1 e^{ix}-ic_2 e^{-ix})+z^{''}(c_1 e^{ix}+c_2 e^{-ix})$$
In order RHS be form $c_3 e^{ix}+c_4 e^{-ix}$,
$$y= (c_3 e^{ix}+c_４ e^{-ix})x$$

 

[arhzz]

Multiplicity is multiplicity of the roots. If for example the characteristic polynomial is $(\lambda -i)^2(\lambda+i)$ then $i$ is a root of the polynomial of degree two since that's the power on $(\lambda-i)$, and $-i$ would have multiplicity 1.

In your first polynomial the roots have multiplicity two, in your second example they have multiplicity 1 (try factoring it)

Hmmm,okay so the trick is multiplicity.But how do I check for multiplicity? By factorization? The example you gave,okay i is the root of the polynom $(\lambda -i)^2$ but how do we know that the multiplicity of -i is 1?

 

[Mark44]

Hmmm,okay so the trick is multiplicity. But how do I check for multiplicity? By factorization? The example you gave,okay i is the root of the polynom $(\lambda -i)^2$ but how do we know that the multiplicity of -i is 1?

From your characteristic equation, which is $\lambda^4 + 2\lambda^2 + 1 = 0$. In factored form this is $(\lambda^2 + 1)^2 = 0$. So both $i$ and $-i$ have multiplicity 2, which means that your solution for this homogeneous problem is going to have to include four functions, which the Wolframalpha solution does.

 

[arhzz]

From your characteristic equation, which is $\lambda^4 + 2\lambda^2 + 1 = 0$. In factored form this is $(\lambda^2 + 1)^2 = 0$. So both $i$ and $-i$ have multiplicity 2, which means that your solution for this homogeneous problem is going to have to include four functions, which the Wolframalpha solution does.

Okay so if I got it right,the power actually determines what the multiplicity is going to be.Kind of hard to see if you don't factor it,I did it like this.

$$ \lambda^4 + 2\lambda^2 + 1 = 0 $$ I replaced the lambda squared with u; $u = \lambda^2$ and than I solved the quadratic equation,and simply reverted the substitution.Guess I need to try to factorize more.

 

[Delta2]

I think there is a theorem in $\mathbb{C}$ that says that the sum of multiplicities of the roots of a polynomial, equals the degree of the polynomial. In this example the degree is 4, so the sum of the multiplicities should equal 4, there are two roots so it is either 2 and 2 or 3 and 1.

 

[Office_Shredder]

Yeah a good gut check is the number of unknowns should be equal to the order of the differential equation.If you really don't want to factor that polynomial, you could just start plugging in things like $xe^{ix}$ and seeing if you find zeros or not, but I doubt that's an easier way to find which roots have multiplicity > 1

 

[Mark44]

Okay so if I got it right,the power actually determines what the multiplicity is going to be.Kind of hard to see if you don't factor it,I did it like this.

$$ \lambda^4 + 2\lambda^2 + 1 = 0 $$ I replaced the lambda squared with u; $u = \lambda^2$ and than I solved the quadratic equation,and simply reverted the substitution.Guess I need to try to factorize more.

With your substitution, you have $u^2 + 2u + 1 = 0$, or $(u + 1)^2 = 0$. Undoing the substitution, you get $(\lambda^2 + 1)^2 = 0$, which is a fourth-degree equation with roots $\lambda = i$ and $\lambda = -1$, both of multiplicity 2.

The order of your DE is four, which means the solution will need four linearly independent basis functions.

 

[anuttarasammyak]

Further to post #3 in this line, the solution of
$$(D^2+1)^n y=0$$
is
$$y=(c_{n-1} x^{n-1} +c_{n-2} x^{n-2}+ ... + c_1 x + c_0)e^{ix}+(d_{n-1} x^{n-1} +d_{n-2} x^{n-2}+ ... + d_1 x + d_0)e^{-ix}$$

 

[arhzz]

With your substitution, you have $u^2 + 2u + 1 = 0$, or $(u + 1)^2 = 0$. Undoing the substitution, you get $(\lambda^2 + 1)^2 = 0$, which is a fourth-degree equation with roots $\lambda = i$ and $\lambda = -1$, both of multiplicity 2.

The order of your DE is four, which means the solution will need four linearly independent basis functions.

Yea I've calculated it all the way through and it was right,the confusion is all cleared up now. Thank you (all of you!)