Q27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 21-28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion about’s x=0 of a general solution to the given differential equation.

(1-x²) y"-y'+y=tan x

Step-by-Step Solution

Answer

The first four nonzero terms in a power series expansion to the given differential equation (1-x²) y"-y'+y=tan x are y(x)=a₀ (1-x²/2-x³/6-x⁴/12-...)+a₁ (x+x²/2+x⁴/24+...)+(x³/6+x⁴/12+...)..

Define power series expansion:

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

Find the expression:

Given,

(1-x²) y"-y'+y=tan x 

The standard form is 

y"-1/(1-x²)y'+1/(1-x²)y=tan x/(1+x²)

Use the formula,

y(x)=Σ_n=0^∞ a_nxⁿ

Taking derivative of the above equation,

y'(x)=Σ_n=1^∞ na_nx^n-1

y"(x)=Σ_n=2^∞ n(n-1)a_nx^n-2

The series expansion is

tan x=x+x³/3+2x⁵/15+...

Substitute the values in the above formula you get,

(1+x²)Σ_n=2^∞ n(n-1)a_nx^n-2 -Σ_n=1^∞ na_nx^n-1 + Σ_n=0^∞ a_nxⁿ =x+x³/3+2x⁵/15+... 

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 =x+x³/3+2x⁵/15+...  

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 n-1=k therefore n=k+1 in the other term.

Σ_k=0^∞ (k+2)(k+1) a_k+2 x^k -Σ_k=2^∞ k(k-1) a_k x^k - Σ_k=0^∞ (k+1) a_k+1 x^k + Σ_k=0^∞ a_k x^k =x+x³/3+2x⁵/15+...   

Hence the expression is Σ_k=0^∞ (k+2)(k+1) a_k+2 x^k -Σ_k=2^∞ k(k-1) a_k x^k - Σ_k=0^∞ (k+1) a_k+1 x^k + Σ_k=0^∞ a_k x^k =x+x³/3+2x⁵/15+... .

Find the first four nonzero terms:

Expand the expression given in the previous step.

(2a₂+a₀-a₁)+(6a₃+a₁-2a₂) x+ (12a₄-2a₂-3a₃+a₂) x²+...=x+x³/3+2x⁵/15+... 

By equating the coefficients you get,

2a₂+a₀-a₁=0

a₂=(a₁-a₀)/2

6a₃+a₁-2a₂=1

a₃=(1-a₀)/6

12a₄-2a₂-3a₃+a₂=0

a₄=a₁/24-a₀/12+1/24

Substitute the coefficient.

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

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

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