# Proof of Differential Equation Identities

• Oct 18th 2009, 11:56 AM
Jiggy
Proof of Differential Equation Identities
Hi guys I'm new here and I thought that maybe you could help me with this problem. I don't quite understand the notation and what the thing is meaning.

Prove the following:

(D-a)(D-b){f(x)e^(λx)} = e^(λx)(D+λ-a)(D+λ-b)f(x)

I just know that $\displaystyle D=\frac {d}{dx}$
But what does the curly parenthesis notation mean?

Any ideas?
• Oct 20th 2009, 09:06 AM
Jester
Quote:

Originally Posted by Jiggy
Hi guys I'm new here and I thought that maybe you could help me with this problem. I don't quite understand the notation and what the thing is meaning.

Prove the following:

(D-a)(D-b){f(x)e^(λx)} = e^(λx)(D+λ-a)(D+λ-b)f(x)

I just know that $\displaystyle D=\frac {d}{dx}$
But what does the curly parenthesis notation mean?

Any ideas?

Expand (the $\displaystyle \{ \}$ are just brackets to group things together)

$\displaystyle (D-a) \left\{ e^{\lambda x} f(x)\right\} = D \left\{e^{\lambda x} f(x)\right\} - a e^{\lambda x} f(x)$

$\displaystyle = D \left(e^{\lambda x}\right)\cdot f(x) + e^{\lambda x} D f(x) - a e^{\lambda x} f(x)$

$\displaystyle = \lambda e^{\lambda x} \cdot f(x) + e^{\lambda x} D f(x) - a e^{\lambda x} f(x)$

then factor

$\displaystyle = e^{\lambda x} \left( D f(x) + (\lambda - a) f(x) \right)$

$\displaystyle = e^{\lambda x} \left( D + \lambda - a \right) f(x).$

The rest is similar.