An ODE involving delta function

[AlonsoMcLaren]

x''+2x'+x=t+delta(t) x(0)=0 x'(0)=1

The textbook, "Elementary differential equations" by Edwards and Penney, gives the answer as -2+t+2exp(-t)+3t exp(-t)

It is clearly wrong, as in this case x'(0)=2, not x'(0)=1.

[nucl34rgg]

I got -2u(t)+t+2exp(-t)+3t exp(-t), with u(t) being the Heaviside function. I got this using a Laplace transform and then doing partial fractions. I think they just left off the u(t) for their solution on accident.

[AlonsoMcLaren]

What do mean by saying that x'(0)=1、 Do we mean x'(0+)=1 or x'(0-)=1?

[nucl34rgg]

EDIT: Whoa, I was way off for my reasoning for x'(0)! Sorry about that. I'd suspect it would be x'(0-) since that is what you use when you take the transform.

[AlonsoMcLaren]

But there is a delta(t) in the input so x'(0-) and x'(0+) cannot be the same. x'(0+)=x'(0-)+1

[hunt_mat]

Why not take Laplace transforms?