Solve the given initial-value problem. \frac{d^2x}{dt^2}=\omega^2 x=F_0\cos(\gamma t),\ x(0)=0,\ x'(0)=0

ajedrezlaproa6j

ajedrezlaproa6j

Answered question

2021-12-30

Solve the given initial-value problem.
d2xdt2=ω2x=F0cos(γt), x(0)=0, x(0)=0

Answer & Explanation

Timothy Wolff

Timothy Wolff

Beginner2021-12-31Added 26 answers

Find the homogeneous solution of the given IVP as follows.
The given IVP is as follows,
d2xdt2=ω2x=F0cos(γt), x(0)=0, x(0)=0 and ωγ
x+ω2x=F0cos(γt), x(0)=x(0)=0 and ωγ
Consider the homogeneous part x+ω2x=0
The characterice equation is m2+ω2=0
m2+ω2=0
⇒=ω
⇒=ωi,ωi
The homogeneous solution is xh=C1cos(ωt)+C2sin(ωt)
Find the particular solution of the given IVP as follows.
The IVP is x+ω2x=F0cos(γt), x(0)=x(0)=0 and ωγ
Let the particular solution be of the form, xp=Acos(γt)
x=γAsin(γt)
x=γ2cos(γt)
γ2cos(γt)+ω2Acos(γt)=F0cos(γt)
Aω2γ2=F0
A=F0+γ2ω2,ω0
The particular solution is xp=F0+γ2ω2cos(γt)
The general solution is of the given IVP is as follows.
The general solution is
x(t)=xh+xp
=C1cos(ωt)+C2sin(ωt)+F0+γ2ω2cos(γt)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?