Paula Boyer

2022-04-22

Particular solution of a system of second order ODE

$\left[\begin{array}{ccc}-k& k& \\ k& -k& \end{array}\right]\left[\begin{array}{c}{x}_{1}(t)\\ {x}_{2}(t)\end{array}\right]+\left[\begin{array}{c}F-k\times a\\ k\times a\end{array}\right]$

$=\frac{{d}^{2}}{d{t}^{2}}\left[\begin{array}{c}{x}_{1}(t)\\ {x}_{2}(t)\end{array}\right]$

gonzakunti2

Beginner2022-04-23Added 16 answers

Step 1

I will write your system as

$x{}^{\u2033}=Mx+b,$

where$x\in {\mathbb{R}}^{2},$ ,

$M=\left(\begin{array}{ccc}-k& k& \\ k& -k& \end{array}\right)$ ,

and

$b=\left(\begin{array}{c}F-k\times a\\ k\times a\end{array}\right)$

assuming that$\times$ is multiplication of scalars.

Now do a linear transformation to$y=x-{M}^{-1}b$ , giving that

$y{}^{\u2033}=My.$

Then I believe this equation has solutions of the form

$y\left(t\right)={y}_{1}\mathrm{exp}\left(Nt\right)+{y}_{2}\mathrm{exp}(-Nt),$

where$y}_{i}\in {\mathbb{R}}^{2},\text{}\mathrm{exp$ is the matrix exponential, and N is any matrix square root of M, satisfying ${N}^{2}=M$

It suffices to find square roots of your M, which may or may not exist, I didn't check.

Edit; I realised that your matrix is not invertible, so my solution doesn't work. I will leave it up in case this solution for invertible M is still useful for someone.

I will write your system as

where

and

assuming that

Now do a linear transformation to

Then I believe this equation has solutions of the form

where

It suffices to find square roots of your M, which may or may not exist, I didn't check.

Edit; I realised that your matrix is not invertible, so my solution doesn't work. I will leave it up in case this solution for invertible M is still useful for someone.

Annie Levine

Beginner2022-04-24Added 15 answers

Step 1

Writing your system as

$x{}^{\u2033}=k(y-x)+F-ak,$

and

$y{}^{\u2033}=k(x-y)+ak,$

leads me to change coordinates to$u=x+y,\text{}v=x-y$ , in which case your equations decouple to

$u{}^{\u2033}=F$

and

$v{}^{\u2033}=-2kv+F-2ak.$

I would expect both of these ODEs to be covered in a first course on differential equations. You can use the linear transformation that I used in my other answer to remove the constant term from the second equation.

Writing your system as

and

leads me to change coordinates to

and

I would expect both of these ODEs to be covered in a first course on differential equations. You can use the linear transformation that I used in my other answer to remove the constant term from the second equation.

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