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

Given,

y"-xy'+2y=cosx

Use the formula,

²â(³æ)=Σ_n=0^∞a_nxⁿ

Taking derivative of the above equation,

²â'(³æ)=Σ_n=1^∞na_nx^n-1

²â"(³æ)=Σ_n=2^∞n(n-1)a_nx^n-2

The series expansion is,

cos x=1-x²/2!+x⁴/4!

Substitute the values in the above formula you get,

Σ_n=2^∞n(n-1)a_nx^n-2 -x Σ_n=1^∞na_nx^n-1 + 2 Σ_n=0^∞a_nxⁿ =1-x²/2!+x⁴/4! -...

(2a₂+a₀)+Σ_k=1^∞ [(k+2)(k+1)a_k+2+(2-k)a_k] x^k =1-x²/2!+x⁴/4! -...

Hence, the expression is (2a₂+a₀)+Σ_k=1^∞ [(k+2)(k+1)a_k+2+(2-k)a_k] x^k =1-x²/2!+x⁴/4! -... .

Expand the expression given in the previous step.

(2a₂+2a₀)+(6a₃+a₁) x+(12a₄)x²+(20a₅-a₃)+.....=1-x²/2!+x⁴/4!

By equating the coefficients you get,

2a₂+2a₀=1

a₂=1/2-a₀

6a₃+a₁=0

a₃=-a₁/6

12a₄= -1/2

a₄= -1/24

20a₅-a₃=0

a₅=-a₁/120

Substitute the coefficient.

y(x)=a₀+a₁x+(1/2-a₀) x²-a₁/6 x³-1/24 x⁴-a₁/120x⁵+...

Hence, the first four nonzero terms is y(x)=a₀+a₁x+(1/2-a₀) x²-a₁/6 x³-1/24 x⁴-a₁/120x⁵+...

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