Posted by **Anonymous** on Wednesday, March 16, 2011 at 1:54pm.

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))

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