StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q24E

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 abouts x=0**** of a general solution to the given differential equation.**

**y"2xy'+3y=x ^{2}**

The first four non-zero terms in a power series expansion to the given differential equation y"2xy'+3y=x^{2} are y(x)=a_{0} (1-3x^{2}/2-x^{4}/8+...)+a_{1} (x-x^{3}/6+...) + (x^{4}/12+...).

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"2xy'+3y=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}

Substitute the values in the above formula you get,

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

In order to make the exponent common for all terms you 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+2 }x^{k}-2Σ_{k}_{=1}^{∞ }k a_{k }x^{k}+3 Σ_{k}_{=0}^{∞ } a_{k }x^{k }=x^{2}

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

Hence the expression is 2a_{2}+3a_{0}+Σ_{k}_{=1}^{∞} [(k+2)(k+1)a_{k+2}+(3-2k)a_{k}] x^{k} =x^{2}.

Expand the expression given in the previous step.

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

By equating the coefficients you get,

2a_{2}+3a_{0}=0

a_{2}= -3a_{0}/2

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

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

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

a_{4}=1/12-a_{0}/8

Substitute the coefficient.

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

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

Hence, the first four non-zero terms is y(x)=a_{0} (1-3x^{2}/2-x^{4}/8+...)+a_{1} (x-x^{3}/6+...) + (x^{4}/12+...).

94% of StudySmarter users get better grades.

Sign up for free