- #1

- 90

- 2

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.

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- Thread starter AlonsoMcLaren
- Start date

- #1

- 90

- 2

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.

- #2

- 149

- 14

- #3

- 90

- 2

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

- #4

- 149

- 14

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.

Last edited:

- #5

- 90

- 2

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

- #6

hunt_mat

Homework Helper

- 1,760

- 27

Why not take Laplace transforms?

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