- #1

- 603

- 6

## Homework Statement

## Homework Equations

DifEqs

## The Attempt at a Solution

y ' = 4C

_{1}e

^{-4x}SinX - 4C

_{2}e

^{-4x}CosX

y'(0) = -1

-1 = 0 - 4C

_{2}

Therefore

C

_{2}= 1/4

Not correct. What am I doing wrong?

- Thread starter Feodalherren
- Start date

- #1

- 603

- 6

DifEqs

y ' = 4C

y'(0) = -1

-1 = 0 - 4C

Therefore

C

Not correct. What am I doing wrong?

- #2

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 14,883

- 1,461

You didn't differentiate correctly. You have to use the product rule.

- #3

- 603

- 6

Sigh... obviously. Can't believe I just did that. Thanks!

- #4

Svein

Science Advisor

- 2,129

- 687

What am I doing wrong?

- [itex] y=c_{1}e^{-4x}cos(x)+c_{2}e^{-4x}sin(x)[/itex]
- [itex]y'=c_{1}(-4e^{-4x}cos(x)-e^{-4x}sin(x))+c_{2}(-4e^{-4x}sin(x)+e^{-4x}cos(x)) [/itex]

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 962

Then, by the product rule, [itex]y'= -4e^{-4x}(C_1 cos(x)+ C_2 sin(x))+ e^{-4x}(-C_1 sin(x)+ C_2 cos(x))[/itex].

- #6

- 603

- 6

Yeah I got it dudes. I was just being stupid and completely forgot the product rule.

Thanks

Thanks

- Last Post

- Replies
- 5

- Views
- 856

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 4

- Views
- 3K

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 2

- Views
- 13K

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 3K

- Replies
- 5

- Views
- 1K