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Q28E

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. **

**y"-(sin x)y=cos x**

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

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,

y"-(sin x)y=cos x

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,

sin x=x-x^{3}/3!+x^{5}/5!-...

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

Substitute the values in the above formula you get,

Σ_{n=2}^{∞ }n(n-1)a_{n}x^{n-2} -(x-x^{3}/3!+x^{5}/5!-...) Σ_{n=0}^{∞ }a_{n}x^{n}

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

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

Hence, the expression is (2a_{2}+6a_{3}x+12a_{4}x^{2}+20a_{5}x^{3}+...) - (x-x^{3}/3!+x^{5}/5!-...) (a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+...) 1-x^{2}/2!+x^{4}/4!-... .

Expand the expression given in the previous step.

(2a_{2}+6a_{3}x+12a_{4}x^{2}+20a_{5}x^{3}+...) - (a_{0}x^{2}+a_{1}x+a_{2}x^{3}+a_{3}x^{4}+...) + (a_{0}x^{3}/3!+a_{1}x^{4}/3!+a_{2}x^{5}/3!+a_{3}x^{6}/3!+...) - (a_{0}x^{5}/5!+a_{1}x^{6}/5!+a_{2}x^{7}/5!+a_{3}x^{8}/5!+...) + ... =1-x^{2}/2!+x^{4}/4!+...

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

By equating the coefficients you get,

2a_{2}=1

a_{2}=1/2

6a_{3}-a_{0}=0

a_{3}=a_{0}/6

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

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

Substitute the coefficient.

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

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

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