Proof of Differential Equation Identities

**Jiggy** · October 18th 2009, 11:56 AM

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)
(D^2+aD+b){f(x)e^(λx)} = e^(λx){(D+λ)^2+a(D+λ)+b}f(x)

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

Any ideas?

**Danny** · October 20th 2009, 09:06 AM

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)
(D^2+aD+b){f(x)e^(λx)} = e^(λx){(D+λ)^2+a(D+λ)+b}f(x)

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

Any ideas?

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

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

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

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

then factor

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

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

The rest is similar.

Thread: Proof of Differential Equation Identities

LinkBack

Thread Tools

Display

Proof of Differential Equation Identities

Similar Math Help Forum Discussions

Valid way of proving vector differential identities

Using suffix notation to prove vector differential identities

Gompertz differential equation proof

Vector differential identities

Differential Equation Proof

Search Tags