Q21E - 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 x=0 of a general solution to the given differential equation. 

y'-xy=sinx

Step-by-Step Solution

Answer

The first four nonzero terms in a power series expansion to the given differential equation y'-xy=sinx are y(x)=a₀ (1+x²/2+x⁴/8+x⁶/48+...)+(x²/2+x⁴/12+11x⁶/720+...).

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,

y'+xy=sinx

Use the formula,

y(x)=Σ_n=0^∞ a_nxⁿ

Taking derivative of the above equation,

y'(x)=Σ_n=0^∞ n a_nx^n-1

The Maclaurin series for sinx is 

sinx=x-x³/3!+x⁵/5! - ...

Substitute the values in the above formula you get,

Σ_n=0^∞ n a_nx^n-1 -x Σ_n=0^∞ a_nxⁿ = x-x³/3! + x⁵/5! - ...

Σ_n=0^∞ n a_nx^n-1 - Σ_n=0^∞ a_nxⁿ⁺¹ = x-x³/3! + x⁵/5! - ... 

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

Σ_k=0^∞ (k+1) a_k+1 x^k - Σ_k=0^∞ a_k-1 x^k= x-x³/3! + x⁵/5! - ... 

a₁+Σ_k=1^∞ (k+1) a_k+1 x^k - Σ_k=1^∞ a_k-1 x^k= x-x³/3! + x⁵/5! - ... 

a₁+Σ_k=1^∞ [(k+1) a_k+1 - a_k-1 ] x^k= x-x³/3! + x⁵/5! - ... 

Hence, the expression is a₁+Σ_k=1^∞ [(k+1) a_k+1 - a_k-1 ] x^k= x-x³/3! + x⁵/5! - ... .

Find the first four nonzero terms:

Expand the expression given in the previous step.

a₁+(2a₂-a₀)x+(3a₃-a₁)x²+(4a₄-a₂)x³+(5a₅-a₃)x⁴+(6a₆-a₄) x⁵+.....=x-x³/6+x⁵/120-x⁷/5040+.....

By equating the coefficients you get,

a₁=0

2a₂-a₀=1

a₂ = (1+a₀)/2

3a₃-a₁=0

3a₃-0=0

a₃=0

4a₄-a₂=-1/6

a₄=1/12+a₀/8

5a₅-a₃=0

a₅=0

6a₆-a₄=1/120

a₆=11/720+a₀/48

Substitute the coefficient.

y(x)=a₀+(1+a₀)/2 x²+ (1/12+a₀/8) x⁴=(11/720+a₀/48) x⁶+...

y(x)=a₀+(1+x²/2+x⁴/8+x⁶/48 +...) + (x²/2+x⁴/12+11x⁶/720 +...) 

Hence the first four nonzero terms are,

y(x)=a₀+(1+x²/2+x⁴/8+x⁶/48 +...) + (x²/2+x⁴/12+11x⁶/720 +...)