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

Haven 2021-01-05 Answered

Verify by direct substitution that the given power series is a solution of the indicated differential equation.

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## Expert Answer

brawnyN
Answered 2021-01-06 Author has 91 answers

Step 1

To verify by direct substitution that the given 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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