Verify by direct substitution that thegiven power series is a solution of the indicated differentialequation.

Question

Verify by direct substitution that the given power series is a solution of the indicated differential equation. $[H i n t : For a power x^{2 n + 1} let k = n + 1.]$

$y = \sum_{(n = 0)}^{\infty} \frac{(- 1)}{2^{2 n} (n!)^{2}} x^{2 n}, x y^{n + y^{'}} = x y = 0$

Answer 1

Step 1

To verify by direct substitution that the given power series: $y = \sum_{(n = 0)}^{\infty} \frac{(- 1)}{2^{2 n} (n!)^{2}} x^{2 n} is a solution of x y^{″} + y^{'} + x y = 0$ .

Step 2

$x y^{″} + y^{'} + x y = \sum_{n = 0}^{\infty} \frac{- 1^{n} 2 n (2 n - 1)}{2^{2 n} (n!)^{2}} x^{2 n - 1}$

$+ \sum_{n = 0}^{\infty} \frac{(- 1)^{n} 2 n}{2^{2 n} (n!)^{2}} x^{2 n - 1}$

$+ \sum_{n = 0}^{\infty} (- 1)^{n} 2^{2 n} (n!)^{2} x^{2 n + 1}$

Now let us substitute $k = n + 1 for the power x^{2 n + 1}$

$x y^{″} + y^{'} + x y = \sum_{k = 1}^{\infty} [\frac{(- 1)^{k} 2 k (2 n - 1)}{2^{2 k} (k!)^{2}} + \frac{(- 1)^{k} 2 k}{2^{2 k} (k!)^{2}} + \frac{(- 1)^{k - 1}}{2^{2 k - 2} ((k - 1)!)^{2}}] x^{2 k - 1}$
$x y^{″} + y^{'} + x y = \sum_{k = 1}^{\infty} [\frac{(- 1)^{k} 2 k}{2^{2 k} (k!)^{2}} - \frac{(- 1)^{k} 2 k}{2^{(} 2 k - 2) ((k - 1)!)^{2}}] x^{2 k - 1}$

Thus, $x y^{″} + y^{'} + x y = 0$ .

Hence, $y = \sum_{n = 0}^{\infty} \frac{(- 1)}{2^{2 n} (n!)^{2}} x^{2 n} is a solution of x y^{″} + y^{'} + x y = 0$ .

Verify by direct substitution that thegiven power series is a solution of the indicated differentialequation.

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