Solving a Diff. Eq. with Orthogonal Vectors

[kq6up]

I am working on a diff eq. that the prof. did as an example in class.

$y^{\prime\prime}-3y^{\prime}+2y=x^2+x+3$

after subbing in I get:

$2a_2-6a_2x-3a_1+2a_2x^2+2a_1x+2a_o=x^2+x+3$

She set aside the $x^2$ terms and set them equal to zero like such:

$2a_2x^2-x^2=0$

I imagine this ok because each order of exponent can be treated like a orthogonal vector in function space? That is why it is ok to pull these out of the original equation? If I had two vectors that I say have to be equal to zero (these for example):

$\mathbf{A}+\mathbf{B}=0$, so $a_x\hat{x}+a_y\hat{y}+b_x\hat{x}+b_y\hat{y}=0$ can be simplified to $a_x=-b_x$ and $a_y=-b_y$.

Are these two examples analogous? If so, I am satisfied that it works. (not saying it doesn't -- I just want to understand things on a deep level).

Thanks,
Chris

 

[LCKurtz]

I am working on a diff eq. that the prof. did as an example in class.

$y^{\prime\prime}-3y^{\prime}+2y=x^2+x+3$

after subbing in I get:

$2a_2-6a_2x-3a_1+2a_2x^2+2a_1x+2a_o=x^2+x+3$

She set aside the $x^2$ terms and set them equal to zero like such:

$2a_2x^2-x^2=0$

I imagine this ok because each order of exponent can be treated like a orthogonal vector in function space? That is why it is ok to pull these out of the original equation? If I had two vectors that I say have to be equal to zero (these for example):

$\mathbf{A}+\mathbf{B}=0$, so $a_x\hat{x}+a_y\hat{y}+b_x\hat{x}+b_y\hat{y}=0$ can be simplified to $a_x=-b_x$ and $a_y=-b_y$.

Are these two examples analogous? If so, I am satisfied that it works. (not saying it doesn't -- I just want to understand things on a deep level).

Thanks,
Chris

You have the right idea. If two polynomials are equal, their corresponding coefficients must be equal, and that is what you are using. You can prove that using derivatives. For example, suppose$$
Ax^2+Bx+C = ax^2+bx+c$$Putting $x=0$ tells you $C=c$. Now take the derivative:$$
2Ax +B = 2ax + b$$Put $x=0$ in that giving $B=b$. Differentiate again and you will see $A=a$. This argument works for higher degress; you just keep differentiating.

 

[kq6up]

Thanks, now it is even more clear to me. That makes perfect sense.

Chris

 

[ehild]

I am working on a diff eq. that the prof. did as an example in class.

$y^{\prime\prime}-3y^{\prime}+2y=x^2+x+3$

after subbing in I get:

$2a_2-6a_2x-3a_1+2a_2x^2+2a_1x+2a_o=x^2+x+3$

Subbing in what for what and where?

She set aside the $x^2$ terms and set them equal to zero like such:

$2a_2x^2-x^2=0$

I imagine this ok because each order of exponent can be treated like a orthogonal vector in function space? That is why it is ok to pull these out of the original equation? If I had two vectors that I say have to be equal to zero (these for example):

$\mathbf{A}+\mathbf{B}=0$, so $a_x\hat{x}+a_y\hat{y}+b_x\hat{x}+b_y\hat{y}=0$ can be simplified to $a_x=-b_x$ and $a_y=-b_y$.

Are these two examples analogous? If so, I am satisfied that it works. (not saying it doesn't -- I just want to understand things on a deep level).

Thanks,
Chris

It is right, the different powers of x can be considered as orthogonal vectors. Two polynomials are identical if the coefficients of all powers of x are the same in both. ehild

 

[kq6up]

Thanks,
Chris