mx''(t)=-kx(t)-cx'(t) I was wondering if this is the right way of

Tayyib Cornish

Tayyib Cornish

Answered question

2022-02-15

mx(t)=kx(t)cx(t)
I was wondering if this is the right way of thinking of it..
1.cx(t)=kx(t)mx(t)
2.integral of kx(t)=cx(t)mx(t)
3.derivative of mx(t)=kx(t)cx(t)
If so, how would I solve 23?

Answer & Explanation

Anderson Higgs

Anderson Higgs

Beginner2022-02-16Added 5 answers

The idea is to use x(t)=(x(t)). So instead of having one equation of second order, you introduce a new function, say y(t)=x(t), and you have two equations of first order:
x(t)=y(t)
y(t)=kmx(t)cmy(t)
Andrew Fenton

Andrew Fenton

Beginner2022-02-17Added 6 answers

To change equations to first order we use substitution.
ϕ1(t)=x(t)
ϕ2(t)=x(t)
Then:
ϕ1(t)=x(t)=ϕ2(t)
ϕ2(t)=x(t)=kϕ1(t)cϕ2(t)m
Now you have a system of two first-order differential equations.

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