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Q25E

Expert-verifiedFound in: Page 450

Book edition
9th

Author(s)
R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages
616 pages

ISBN
9780321977069

**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+x ^{2})y"-xy'+y=e**

The first four nonzero terms in a power series expansion to the given differential equation (1+x^{2})y"-xy'+y=e^{-x} is y(x)=a_{0} (1-x^{2}/2+...)+a_{1} (x-x^{4}/12+...) + (x^{2}/2-x^{3}/6+x^{4}/24+...).

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_{n}x^{n}

Given,

(1+x^{2})y"-xy'+y=e^{-x}

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

y"-x/1+x^{2} y'+1/1+x^{2} y=e^{-x}/1+x^{2}

Use the formula,

y(x)=Σ_{n=0}^{∞ }a_{n}x^{n}

Taking derivative of the above equation,

y'(x)=Σ_{n=1}^{∞ }na_{n}x^{n-1}

y"(x)=Σ_{n=2}^{∞ }n(n-1)a_{n}x^{n-2}

The series expansion is,

e^{-x}=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

Substitute the values in the above formula you get,

(1+x^{2})Σ_{n=2}^{∞ }n(n-1)a_{n}x^{n-2} +x Σ_{n=1}^{∞ }na_{n}x^{n-1} + Σ_{n=0}^{∞ }a_{n}x^{n} =1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+.....

Σ_{n=2}^{∞ }n(n-1)a_{n}x^{n-2}+Σ_{n=2}^{∞ }n(n-1)a_{n}x^{n}

-Σ_{n=1}^{∞ }na_{n}x^{n-1} + Σ_{n=0}^{∞ }a_{n}x^{n} =1-x+x^{2}/2!-x^{3}/3!+x^{4}/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_{k}x^{k}+ Σ_{k=2}^{∞} k(k-1)a_{k}x^{k} - Σ_{k=1}^{∞} ka_{k}x^{k}

=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

(2a_{2}+a_{0}+6a_{3}+a_{1}x-a_{1}x)+Σ_{k=2}^{∞} ((k+2) (k+1)a_{k+2 }+k(k-1)a_{k}-ka_{k}+a_{k}) x^{k}

=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

(2a_{2}+a_{0}+6a_{3})+Σ_{k=2}^{∞} ((k+2) (k+1)a_{k+2 }+(k^{2}-2k+1)a_{k}) x^{k}

=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

(2a_{2}+a_{0}+6a_{3})+Σ_{k=2}^{∞} ((k+2) (k+1)a_{k+2 }+(k-1)^{2}a_{k}) x^{k}

=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

Hence the expression is (2a_{2}+a_{0}+6a_{3})+Σ_{k=2}^{∞} ((k+2) (k+1)a_{k+2 }+(k-1)^{2}a_{k}) x^{k} =1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+....

Expand the expression given in the previous step.

(2a_{2}+a_{0}+6a_{3})+(12a_{4}+a_{1}) x^{2}+(20a_{5}+4a_{3}) x^{2}+...=1-x+x^{2}/2!-x^{3}/3!+x^{4}/4!+...

By equating the coefficients you get,

2a_{2}+a_{0}=1

a_{2}=(1-a_{0})/2

6a_{3 }= -1

a_{3}= -1/6

12a_{4}+a_{1}=1/2

a_{4}=1/24-a_{1}/12

Substitute the coefficient.

y(x)=a_{0}+a_{1}x+((1-a_{0})/2)x^{2}+(-1/6)x^{3}+(1/24-a_{1}/12) x^{4}+...

=a_{0} (1-x^{2}/2+...)+a_{1} (x-x^{4}/12+...) + (x^{2}/2-x^{3}/6+x^{4}/24+...)

Hence, the first four nonzero terms is y(x)=a_{0} (1-x^{2}/2+...)+a_{1} (x-x^{4}/12+...) + (x^{2}/2-x^{3}/6+x^{4}/24+...).

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