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. [Hint:For a power x2n+1 let k=n+1.]

y=(n=0)(1)22n(n!)2x2n, xyn+y=xy=0

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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:y=(n=0)(1)22n(n!)2x2n is a solution of xy+y+xy=0.

Step 2

xy+y+xy=n=01n2n(2n1)22n(n!)2x2n1

+n=0(1)n2n22n(n!)2x2n1

+n=0(1)n22n(n!)2x2n+1

Now let us substitute k=n+1for the power x2n+1

 xy+y+xy=k=1[(1)k2k(2n1)22k(k!)2+(1)k2k22k(k!)2+(1)k122k2((k1)!)2]x2k1
xy+y+xy=k=1[(1)k2k22k(k!)2(1)k2k2(2k2)((k1)!)2]x2k1

Thus, xy+y+xy=0.

Hence,y=n=0(1)22n(n!)2x2n is a solution of xy+y+xy=0.

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