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Q26E

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"-xy'+2y=cosx**

The first four non-zero terms in a power series expansion to the given differential equation y"-xy'+2y=cosx are y(x)=a_{0}+a_{1}x+(1/2-a_{0}) x^{2}-a_{1}/6 x^{3}^{ }-1/24 x^{4}-a_{1}/120x^{5}+....

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"-xy'+2y=cosx

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,

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

(2a_{2}+a_{0})+Σ_{k}_{=1}^{∞} [(k+2)(k+1)a_{k+2}+(2-k)a_{k}] x^{k} =1-x^{2}/2!+x^{4}/4! -...

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

Expand the expression given in the previous step.

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

By equating the coefficients you get,

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

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

6a_{3}+a_{1}=0

a_{3}=-a_{1}/6

12a_{4}= -1/2

a_{4}= -1/24

20a_{5}-a_{3}=0

a_{5}=-a_{1}/120

Substitute the coefficient.

y(x)=a_{0}+a_{1}x+(1/2-a_{0}) x^{2}-a_{1}/6 x^{3}^{ }-1/24 x^{4}-a_{1}/120x^{5}+...

Hence, the first four nonzero terms is y(x)=a_{0}+a_{1}x+(1/2-a_{0}) x^{2}-a_{1}/6 x^{3}^{ }-1/24 x^{4}-a_{1}/120x^{5}+...

.

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