Thread: Powers series and diffy q's

f(x) = E(0, inf) [((-1)^n)(x^2n)]/(2n+1)!

And I found f'(x) to be E (0,inf) [2n((-1)^n)(x^(2n-1))]/(2n+1)!

The question is:
Show that y = f(x) is a solution to the differention equation xy'+y=cosx

Not sure really where to start or how to show this. Help appreciated.

Originally Posted by Nerd

f(x) = E(0, inf) [((-1)^n)(x^2n)]/(2n+1)!

And I found f'(x) to be E (0,inf) [2n((-1)^n)(x^(2n-1))]/(2n+1)!

The question is:
Show that y = f(x) is a solution to the differention equation xy'+y=cosx

Not sure really where to start or how to show this. Help appreciated.

Take y=f(x) and differenciate y term-by-term to get y'.
Then add y + y' meaning term-by-term addition.
Show that this sum is equal to the sine series expansion.

I'll try that. Now that I look at it, that doesn't seem too hard. Thanks. Sometimes things come slow to me.

Oh some more questions. Wouldn't I multiply x with y before doin the term by term addition? Why would I use sine instead of cosine? And would the sine expansion be the McLaurin series expansion for sine?

Originally Posted by Nerd

f(x) = E(0, inf) [((-1)^n)(x^2n)]/(2n+1)!

And I found f'(x) to be E (0,inf) [2n((-1)^n)(x^(2n-1))]/(2n+1)!

The question is:
Show that y = f(x) is a solution to the differention equation xy'+y=cosx

Not sure really where to start or how to show this. Help appreciated.

xy' + y = cosx
= x*E(0,inf) 2n(-1)^n*x^(2n-1)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!
= E(0,inf) 2n(-1)^n*x^(2n-1+1)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!
= E(0,inf) 2n(-1)^n*x^(2n)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!}
= E(0,inf) [2n(-1)^n*x^(2n)/(2n + 1)! + (-1)^n*x^(2n)/(2n + 1)!]
= E(0,inf) [(-1)^n*x^(2n)/[(2n)!(2n + 1)]*(2n + 1)]
= E(0,inf) [(-1)^n*x^(2n)/(2n)!] = cosx

Awesome. So I was working in the right direction. Since cosine is a well known expansion, I wouldn't really need to show how cos x = E(0,inf) [(-1)^n*x^(2n)/(2n)!] right? It basically understood to be true?

Originally Posted by Nerd

Awesome. So I was working in the right direction. Since cosine is a well known expansion, I wouldn't really need to show how cos x = E(0,inf) [(-1)^n*x^(2n)/(2n)!] right? It basically understood to be true?

If you know the expansion of cosine then there is no need to show how you got it, assuming your teacher is ok with that.

Originally Posted by Nerd

Awesome. So I was working in the right direction. Since cosine is a well known expansion, I wouldn't really need to show how cos x = E(0,inf) [(-1)^n*x^(2n)/(2n)!] right? It basically understood to be true?

Originally Posted by EcMathGeek

If you know the expansion of cosine then there is no need to show how you got it, assuming your teacher is ok with that.

Strictly mathematically the series for cosine is a definition rather than a theorem.