Haven

2021-01-05

Make sure the provided power series is a solution to the given differential equation by directly substituting it.

brawnyN

Step 1

Direct substitution can be used to confirm that the specified power series:.

Step 2

$x{y}^{″}+{y}^{\prime }+xy=\sum _{n=0}^{\mathrm{\infty }}\frac{-{1}^{n}2n\left(2n-1\right)}{{2}^{2n}\left(n!{\right)}^{2}}{x}^{2n-1}$

$+\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}2n}{{2}^{2n}\left(n!{\right)}^{2}}{x}^{2n-1}$

$+\sum _{n=0}^{\mathrm{\infty }}\left(-1{\right)}^{n}{2}^{2n}\left(n!{\right)}^{2}{x}^{2n+1}$

Now let us substitute

$x{y}^{″}+{y}^{\prime }+xy=\sum _{k=1}^{\mathrm{\infty }}\left[\frac{\left(-1{\right)}^{k}2k\left(2n-1\right)}{{2}^{2k}\left(k!{\right)}^{2}}+\frac{\left(-1{\right)}^{k}2k}{{2}^{2k}\left(k!{\right)}^{2}}+\frac{\left(-1{\right)}^{k-1}}{{2}^{2k-2}\left(\left(k-1\right)!{\right)}^{2}}\right]{x}^{2k-1}$
$x{y}^{″}+{y}^{\prime }+xy=\sum _{k=1}^{\mathrm{\infty }}\left[\frac{\left(-1{\right)}^{k}2k}{{2}^{2k}\left(k!{\right)}^{2}}-\frac{\left(-1{\right)}^{k}2k}{{2}^{\left(}2k-2\right)\left(\left(k-1\right)!{\right)}^{2}}\right]{x}^{2k-1}$

Thus, $x{y}^{″}+{y}^{\prime }+xy=0$.

Hence,.

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