Particular solution of a system of second order ODE[−kkk−k][x1(t)x2(t)]+[F−k×ak×a]=d2dt2[x1(t)x2(t)]

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gonzakunti2 · Accepted Answer

Step 1I will write your system asx″=Mx+b,where x∈R2,,M=(−kkk−k),andb=(F−k×ak×a)assuming that × is multiplication of scalars.Now do a linear transformation to y=x−M−1b, giving thaty″=My.Then I believe this equation has solutions of the formy(t)=y1exp⁡(Nt)+y2exp⁡(−Nt),where yi∈R2, exp⁡ is the matrix exponential, and N is any matrix square root of M, satisfying N2=MIt suffices to find square roots of your M, which may or may not exist, I didn&#039;t check.Edit; I realised that your matrix is not invertible, so my solution doesn&#039;t work. I will leave it up in case this solution for invertible M is still useful for someone.

Annie Levine · Accepted Answer

Step 1Writing your system asx″=k(y−x)+F−ak,andy″=k(x−y)+ak,leads me to change coordinates to u=x+y, v=x−y, in which case your equations decouple tou″=Fandv″=−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.

Particular solution of a system of second order ODE \(\begin{bmatrix}-k &

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