Power Series - Differential Equation (check my answer)

[vucollegeguy]

Using the power series method to solve the differential equation
y'+xy=0 when y(0)=1

Write the solution in the form of a power series and then recognize what function it represents.
************************************

My answer:

\sum(-1)^k*[(x^2k)/(2^k)*k!]

Is my answer correct?
Is it written in the form of a power series?
What function does it represent?

Thanks to anyone that helps!

 

[Dick]

Yes, it's correct. And yes, it's a power series. It looks like exp(f(x)) to me. Can you guess what f(x) might be?

 

[vucollegeguy]

I'd guess and say that f(x)=cos(x).
So the function would be f(x)=x²cos(x).

Please correct me if I am wrong.
Thank you for helping thus far.

 

[HallsofIvy]

Not quite. Your sum reduces to
\sum(-1)^k \frac{\left(\frac{x^2}{2}\right)^2}{k!}
which is a "cosine" but it is
cos(\frac{x}{2})

x^2cos(x)
would be
x^2\sum(-1)^k \frac{x^k}{k!}= \sum (-1)^k \frac{x^{k+2}}{k!}

 

[vucollegeguy]

Ok -got it.
A big "THANKKK YOUUU!" to all for all of your help!