91Ó°ÊÓ

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,

²â(³æ)=Σ_n=0^∞a_nxⁿ

Let,

³æ(³Ù)=Σ_n=0^∞a_ntⁿ

Taking derivative of the above equation,

³æ'(³Ù)=Σ_n=1^∞na_nt^n-1

³æ"(³Ù)=Σ_n=2^∞n(n-1)a_nt^n-2

The Maclaurin series is,

e^-η³Ù=∑_n=0^∞(-η³Ù)/n!

=∑_n=0^∞(η)ⁿ/n! t^η

Replace this in the equation.

∑_n=2^∞ n(n+1)a_nt^n-2+∑_n=0^∞(η)ⁿ/n! t^η ∑_n=0^∞a_n t^η=0

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

2a₂+a₀=0

a₂=-a₀/2

Hence the expression is a₂=-a₀/2.

Now we have to find the coefficients.

a₂=-1/2

6a₃-η²¹₀+a₁=0

a₃=(η²¹₀-a₁)/6

=η/6

(-η)²/2! a₀+(-η)¹/1! a₁+(-η)⁰/0! a₂=η²/2-η²¹₁+a₂

12a₄+η²/2-η²¹₁+a₂=0

a₄= (-η²/2a₀+η a₁-a₂)/12

=1-η²/24

Substitute the value of coefficients in the expression.

x(t)=1-1/2t²+η/6 t³+(1-η²)/24 t⁴+...

Hence the first four terms are x(t)=1-1/2t²+η/6 t³+(1-η²)/24 t⁴+... .

For η =0 the given equation is,

x"+x=0

The solution is

x(t)=cos t

From part (a) the solution will be,

x(t)=1-1/2t²+1/24t⁴+...

Since this is a Taylor’s series for cos t the expansion in part (a) agrees with the expansion in part (b).