Explanation Given: The differential equation x 2 y " + x y ' + ( x 2 − 4 ) y = 0 .... ( 1 ) And y = J 2 ( x ) = ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 2 .... ( 2 ) . Proof: It is given that J 2 ( x ) = ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 2 Now differentiate the equation (2) with respect to x and we get: y ' = J 2 ' ( x ) = ∑ k = 1 ∞ ( − 1 ) k ( 2 k + 2 ) 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 1 = ∑ k = 1 ∞ ( − 1 ) k ( k + 1 ) 2 2 k + 1 k ! ( k + 3 ) ! x 2 k + 1 .... ( 3 ) y ' ' = J 2 ' ( x ) = ∑ k = 2 ∞ ( − 1 ) k ( k + 1 ) ( 2 k + 1 ) 2 2 k + 1 k ! ( k + 3 ) ! x 2 k .... ( 4 ) Now, the differential equation is: x 2 y " + x y ' + ( x 2 − 4 ) y = 0 .... ( 1 ) Now, substitute the values of y, y’ and y’’into the equation (1) and we get: x 2 y " + x y ' + ( x 2 − 4 ) y = x 2 ∑ k = 2 ∞ ( − 1 ) k ( k + 1 ) ( 2 k + 1 ) 2 2 k + 1 k ! ( k + 3 ) ! x 2 k + x ∑ k = 1 ∞ ( − 1 ) k ( k + 1 ) 2 2 k + 1 k ! ( k + 3 ) ! x 2 k + 1 . + ( x 2 − 4 ) ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 2 = ∑ k = 2 ∞ ( − 1 ) k ( k + 1 ) ( 2 k + 1 ) 2 2 k + 1 k ! ( k + 3 ) ! x 2 k + 2 + ∑ k = 1 ∞ ( − 1 ) k ( k + 1 ) 2 2 k + 1 k !

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 11.6, Problem 63E

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$J_{2} (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2^{2 k + 2} k! (k + 3)!} x^{2 k + 2}$ is a solution of the Bessel differential equation of order 2: $x^{2} y " + x y' + (x^{2} - 4) y = 0$ .

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Chapter 11.6, Problem 62E

Chapter 11.6, Problem 64E

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To prove: J 2 ( x ) = ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 2 is a solution of the Bessel differential equation of order 2: x 2 y " + x y ' + ( x 2 − 4 ) y = 0 .

Solution Summary: The author proves that x 2 k + 2 is a solution of the Bessel differential equation of order 2.

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To prove:

$J_{2} (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2^{2 k + 2} k! (k + 3)!} x^{2 k + 2}$ is a solution of the Bessel differential equation of order 2: $x^{2} y " + x y' + (x^{2} - 4) y = 0$ .

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Chapter 11 Solutions

To prove: J 2 ( x ) = ∑ k = 0 ∞ ( − 1 ) k 2 2 k + 2 k ! ( k + 3 ) ! x 2 k + 2 is a solution of the Bessel differential equation of order 2: x 2 y " + x y ' + ( x 2 − 4 ) y = 0 .

Solution Summary: The author proves that x 2 k + 2 is a solution of the Bessel differential equation of order 2.

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To prove: J2(x)=∑k=0∞(−1)k22k+2k!(k+3)!x2k+2 is a solution of the Bessel differential equation of order 2:x2y"+xy'+(x2−4)y=0.

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Chapter 11 Solutions

To prove:

$J_{2} (x) = \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2^{2 k + 2} k! (k + 3)!} x^{2 k + 2}$ is a solution of the Bessel differential equation of order 2: $x^{2} y " + x y' + (x^{2} - 4) y = 0$ .