Q23E - 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.

z"+xz'+z=x²+2x+1

Step-by-Step Solution

Answer

The first four nonzero terms in a power series expansion to the given differential equation z"+xz'+z=x²+2x+1 is,

z(x)=a₀ (1-x²/2+x⁴/8+...)+a₁ (x-x³/3+...) + (x²/2+x³/3-x⁴/24+...)

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,

z"+xz'+z=x²+2x+1

Use the formula,

z(x)=Σ_n=0^∞ a_nxⁿ

Taking derivative of the above equation,

z'(x)=Σ_n=1^∞ na_nx^n-1

z"(x)=Σ_n=2^∞ n(n-1)a_nx^n-2

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 + Σ_n=0^∞ a_nxⁿ =x²+2x+1

In order to make the exponent common for all terms we will make the substitution n-1=k therefore k=n+2 in first term and k=n+2 in the other term.

Σ_k=0^∞ (k+2)(k+1)a_k+2 x^k+ Σ_k=1^∞ k a_k x^k+ Σ_k=0^∞  a_k x^k =x²+2x+1 

2a₂+a₀+Σ_k=1^∞  [(k+2)(k+1)a_k+2+(k+1)a_k] x^k =x²+2x+1  

Hence the expression is 2a₂+a₀+Σ_k=1^∞  [(k+2)(k+1)a_k+2+(k+1)a_k] x^k =x²+2x+1  .

Find the first four nonzero terms:

Expand the expression given in the previous step.

(2a₂+a₀)+(6a₃+2a₁) x+(12a₄+3a₂)x²+.....=x²+2x+1

By equating the coefficients you get,

2a₂+a₀=1

a₂=(1-a₀)/2

6a₃+2a₁=2

a₃=1/3-a₁/3

12a₄+3a₂=1

a₄=a₀/8-1/24

Substitute the coefficient.

z(x)=a₀+a₁x+((1-a₀)/2)x²+((a₁-1)/3)x³+(a₀/8-1/24) x⁴+...

=a₀ (1-x²/2+x⁴/8+...)+a₁ (x-x³/3+...) + (x²/2+x³/3-x⁴/24+...)