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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,

测(虫)=危_n=0^鈭�a_nxⁿ

Given,

mx"(t)+bx'(t)+ke^-畏迟 x(t)=0

Let

虫(迟)=危_n=0^鈭�a_ntⁿ

Taking derivative of the above equation,

虫'(迟)=危_n=1^鈭�na_nt^n-1

虫"(迟)=危_n=2^鈭�n(n-1)a_nt^n-2

The Maclaurin series is,

e^-t = 鈭�_n=0^鈭� (-t)ⁿ/n!

=鈭�_n=0^鈭� (-1)ⁿ tⁿ/n!

Replace this in the equation.

鈭�_n=2^鈭� n(n-1)a_nt^n-2+2鈭�_n=1^鈭� na_nt^n-1 +鈭�_n=0^鈭� (-1)ⁿtⁿ/苍!+鈭�_n=0^鈭�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+....