Question

Differential equations, initial value problem.

The general equation of motion is:
mx"+Bx'+kx=f(t), where the independent variable is t, and the displacement x is the dependent variable.

In this case, external force f(t)=0, so
mx"+Bx'+kx=0
substitute
m=0.25,
k=4,
B=1
we have a differential equation of motion with constant coefficients:
0.25x"+x'+4x=0
Divide by the leading coefficient:
x"+4x'+16x=0
Auxiliary equation:
m²+4m+16=0
m1=-2+2√3, m2=-2-2√3
α=-2, β=2√3
So the solution for x is:
x=e-2t(C1*cos(βt)+C2*sin(βt))
Since x(0)=-0.5, we have
-0.5=1*(C1+C2*0)
or C1=-0.5

Find
x'
=dx/dt
=a*%e^(a*t)*(sin(b*t)*C2+cos(b*t)*C1)+%e^(a*t)*(b*cos(b*t)*C2-b*sin(b*t)*C1)
substitute
x'(0)=-1 to get
-1=αC1+βC2
-1=-2C1+2√(3)C2
C2=(-1+2(-0.5))/2√(3)
=-√(3)/3
Therefore the equation of motion is:
x(t)=e-2t(-0.5cos(βt)-(√(3)/3)sin(βt))

Answer 1

Is this a question on the above response?

(see: http://www.jiskha.com/display.cgi?id=1299811520)