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y'' + xy = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck-1 for k = 1, 2, 3,

y'' + xy = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck-1 for k = 1, 2, 3,

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y'' + xy = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck-1 for k = 1, 2, 3,

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y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,

y'' - y' = 0 Consider the following differential equation to be solved using a power series. Using the substitution y = n = 0 cn x^n, find an expression for ck+2 in terms of ck+1 for k = 0, 1, 2,
Solution
0 0

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn

This is a short reading exercise on reindexing. There is no standard name to this operation. Some people call it shifting the summation. Consider the notation for a series P∞ n=1 an. The symbol n is called the index of summation. The lower bound of the series is 1 . If we write P∞ n=u bn, then the lower bound of the series is u. Reindexing means to change lower bound of the series without changing the series. For example, the series 1 + 1 4 + 1 9 + 1 16 + 1 25 + • • • can be written as P∞ n=1 1 n2 or P∞ n=3 1 (n−2)2 . Please verify this by writing down the first few terms of both series. Therefore, we have an equality between series X∞ n=1 1 n2 = X∞ n=3 1 (n − 2)2 Notice we DO NOT simply replace n = 1 by n = 3. The expression for the terms of the series also have to change for the equality to hold. To reindex by symbolic manipulation, we introduce a change of variable. In this example, we start with the series P∞ n=1 1 n2 . We wish to reindex from n = 1 to n = 3, meaning the lower bound of the series changes from n = 1 to n = 3 without changing the series. We introduce m = n + 2 (because m = 3 when n = 1 ). Isolate for n to get n = m − 2, and we replace every occurrence of n in the series by the expression on the right-hand side (which is m − 2 in this example). The series becomes X∞ n=1 1 n2 = X∞ m−2=1 1 (m − 2)2 ( replace n by m − 2) = X∞ m=3 1 (m − 2)2 ( reorganize the lower bound of summation) = X∞ n=3 1 (n − 2)2 (change m back to n ) In general, if we wish to reindex a series P∞ n=u an from n = u to n = v, we introduce a change of variable m = n+(v−u) and perform similar manipulations as shown above. In the example above, u = 1 and v = 3. Let r be a non-zero constant. You are asked to reindex the series P∞ n=1 1 rn−1 from n = 1 to n = 0. What is the result of reindexing? Select one or more: a. P∞ n=0 1 rn+1 b. P∞ n=0 1 rn−1 c. P∞ n=0 1 rn
Solution
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y'' + x^2y' + xy = 0 Find two power series solutions of the given differential equation about the ordinary point x = 0.
Solution
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