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,

(1+x²)y"-xy'+y=e^-x

^{The above equation is written in the standard form as,}

y"-x/1+x² y'+1/1+x² y=e^-x/1+x²

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,

e^-x=1-x+x²/2!-x³/3!+x⁴/4!+...

Substitute the values in the above formula you get,

(1+x²)Σ_n=2^∞n(n-1)a_nx^n-2 +x Σ_n=1^∞na_nx^n-1 + Σ_n=0^∞a_nxⁿ =1-x+x²/2!-x³/3!+x⁴/4!+.....

Σ_n=2^∞n(n-1)a_nx^n-2+Σ_n=2^∞n(n-1)a_nxⁿ

-Σ_n=1^∞na_nx^n-1 + Σ_n=0^∞a_nxⁿ =1-x+x²/2!-x³/3!+x⁴/4!+...

In order to make the exponent common for all terms we will make the substitution n-2=k therefore k=n+2 in first term and k=n+2 in the other term.

Σ_k=0^∞ (k+2) (k+1)a_kx^k+Σ_k=2^∞ k(k-1)a_kx^k - Σ_k=1^∞ ka_kx^k

=1-x+x²/2!-x³/3!+x⁴/4!+...

(2a₂+a₀+6a₃+a₁x-a₁³æ)+Σ_k=2^∞ ((k+2) (k+1)a_k+2+k(k-1)a_k-ka_k+a_k) x^k

=1-x+x²/2!-x³/3!+x⁴/4!+...

(2a₂+a₀+6a₃)+Σ_k=2^∞ ((k+2) (k+1)a_k+2+(k²-2k+1)a_k) x^k

=1-x+x²/2!-x³/3!+x⁴/4!+...

(2a₂+a₀+6a₃)+Σ_k=2^∞ ((k+2) (k+1)a_k+2+(k-1)²a_k) x^k

=1-x+x²/2!-x³/3!+x⁴/4!+...

Hence the expression is (2a₂+a₀+6a₃)+Σ_k=2^∞ ((k+2) (k+1)a_k+2+(k-1)²a_k) x^k =1-x+x²/2!-x³/3!+x⁴/4!+....

Expand the expression given in the previous step.

(2a₂+a₀+6a₃)+(12a₄+a₁) x²+(20a₅+4a₃) x²+...=1-x+x²/2!-x³/3!+x⁴/4!+...

By equating the coefficients you get,

2a₂+a₀=1

a₂=(1-a₀)/2

6a₃= -1

a₃= -1/6

12a₄+a₁=1/2

a₄=1/24-a₁/12

Substitute the coefficient.

y(x)=a₀+a₁x+((1-a₀)/2)x²+(-1/6)x³+(1/24-a₁/12) x⁴+...

=a₀ (1-x²/2+...)+a₁ (x-x⁴/12+...) + (x²/2-x³/6+x⁴/24+...)

Hence, the first four nonzero terms is y(x)=a₀ (1-x²/2+...)+a₁ (x-x⁴/12+...) + (x²/2-x³/6+x⁴/24+...).