Aging spring. As a spring ages, its “spring constant” decreases on value. One such model for a mass-spring system with an aging spring is

mx"(t)+bx'(t)+ke^-ηt x(t)=0

where mis the mass,the damping constant, kand ηpositive constants and x(t)displacement of the spring from equilibrium position. Let m=1 kg, b=2 N sec/m, k=1 N/m,η =1 sec^-1. The system is set in motion by displacing the mass 1mfrom it equilibrium position and releasing it (x(0)=1, x'(0)=0). Find at least the first four nonzero terms in a power series expansion of about t=0of displacement.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable.

It is generally given by the formula,

y(x)=Σ_n=0^∞a_nxⁿ

Given,

mx"(t)+bx'(t)+ke^-ηt x(t)=0

Let

x(t)=Σ_n=0^∞a_ntⁿ

Taking derivative of the above equation,

x'(t)=Σ_n=1^∞na_nt^n-1

x"(t)=Σ_n=2^∞n(n-1)a_nt^n-2

The Maclaurin series is,

e^-t = ∑_n=0^∞ (-t)ⁿ/n!

=∑_n=0^∞ (-1)ⁿ tⁿ/n!

Replace this in the equation.

∑_n=2^∞ n(n-1)a_nt^n-2+2∑_n=1^∞ na_nt^n-1 +∑_n=0^∞ (-1)ⁿtⁿ/n!+∑_n=0^∞a_ntⁿ=0

You will set coefficients equal to zero. The expression is,

2a₂+2a₁+a₀=0

a₂= - (2a₁+a₀)/2

Hence the expression is a₂= - (2a₁+a₀)/2.

Now you will find the coefficient.

a₂= - (2a₁+a₀)/2

= -[2 (0)+1]/2

=-1/2

6a₃+4a₂-a₀+a₁=0

a₃=(a₀-a₁ -4a₂)/6

= -[4(-1/2)+1-0]/6

=1/2

12a₄+6a₃+1/2a₀-a₁+a₂=0

a₄= (-6a₃-1/2a₀+a₁-a₂)/12

=[-6(1/2)-1/2(1)+0-(-1/2)]/12

= -1/4

Substitute the coefficients in the expression.

x(t)=1-1/2t²+1/2t³-1/4t⁴+...

Hence, the first four nonzero terms are x(t)=1-1/2t²+1/2t³-1/4t⁴+....