Q22E - 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 non-zero terms in a power series expansion about’s x=0 of a general solution to the given differential equation.

w'+xw=e^x

Step-by-Step Solution

Answer

The first four nonzero terms in a power series expansion to the given differential equation w'+xw=e^x are w(x)=a₀ (1-x²/2+x⁴/8+...)+(x+x²/2-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,

w'+xw=e^x

Use the formula,

w(x)=Σ_n=0^∞a_nxⁿ

Taking derivative of the above equation,

w(x)=Σ_n=1^∞na_nx^n-1

The Maclaurin series for e^x is,

e^x=1+x+x²/2!+x³/3!+.....

Substitute the values in the above formula you get,

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

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

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

Σk₌₀^∞ (k+1)a_k+1 x^k + Σk₌₁^∞ a_k-1 x^k =1+x+x²/2!+x³/3!

a₁+Σk₌₁^∞ (k+1)a_k+1 x^k + Σk₌₁^∞ a_k-1 x^k =1+x+x²/2!+x³/3! 

a₁+Σk₌₁^∞ ((k+1)a_k+1 +a_k-1) x^k=1+x+x²/2!+x³/3! 

Hence the expression is a₁+Σk₌₁^∞ ((k+1)a_k+1 +a_k-1) x^k=1+x+x²/2!+x³/3!.

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+...=1+x+x²/2!+x³/3!

By equating the coefficients you get,

a₁=1

2a₂+a₀=1

a₂=(1-a₀)/2

3a₃+a₁=0

a₃=- -1/6

4a₄+a₂=1/6

a₄=-1/12+a₀/8

Substitute the coefficient.

w(x)=a₀+x+(1-a₀)/2 x²-1/6 x³+(-1/12+a₀/8) x⁴+...

=a₀+ (1-x²/2+x⁴/4+...)+ (x+x²/2-x³/6-x⁴/12-...)

Hence the first four nonzero terms are,

w(x)=a₀+ (1-x²/2+x⁴/4+...)+ (x+x²/2-x³/6-x⁴/12-...)