# Trouble solving this differential equation

1. Feb 20, 2004

### BViper

initial conditions: y(0)=0.1
y'(0)=0
30000*y''+1462163*y+100000*y'=A*sin(6.98t)

im trying to get A (the amplitude). thanks for the help

2. Feb 22, 2004

"A" is not uniquely determined by the conditions that you gave. Varying "A" would change the solution to the equation, but the solution would still exist.

Are you looking how to solve y(t)?

3. Feb 22, 2004

### BViper

yeah thats it.. gotta get y(t) before i can get A

4. Feb 22, 2004

No, you don't need to get y(t) before you get A. As I said before, A is not uniquely determined in this problem. The problem does not require that A take a single value or even a finite number of values. A could be anything (okay, almost anything) and y(t) would still exist.

If you'd like to solve for y(t) in terms of A and t, then that can be done. If you'd like to do that, then perhaps you could supply more information regarding what method you want see (perhaps one you've been using recently?).

5. Feb 27, 2004

### Orion1

implicit differentiation...

initial conditions:
y(0)=0.1
y'(0)=0
30000*y''+1462163*y+100000*y'=A*sin(6.98t)

Uncertain if this approach is correct....

Implicit Differentiation:
ay" + by + cy' = A*sin(d*t)

(d/dt)[ay" + by + cy'] = (d/dt)[A*sin(d*t)]

(d/dt)[ay"] + (d/dt)[by] + (d/dt)[cy'] = (d/dt)[A*sin(d*t)]

a(dy"/dt) + b(dy/dt) + c(dy'/dt) = A(d/dt)[sin(d*t)]

a(dy"/dt) + b(dy/dt) + c(dy'/dt) = A*cos(d*t)

Last edited: Feb 27, 2004
6. Mar 4, 2004

### HallsofIvy

Staff Emeritus
Orion1: The only variables mentioned in this problem are y and t.
y' and y" already ARE the derivatives with respect to time. There is nothing gained by differentiating again.

Cookiemonster's point was correct: the initial value problem:
initial conditions:
y(0)=0.1
y'(0)=0
30000*y''+1462163*y+100000*y'=A*sin(6.98t)

Has a unique solution for every possible value of A. It is impossible to "determine A" from what is given.