# Resulting system of equations is not linearly independent

1. Apr 16, 2015

### e^(i Pi)+1=0

1. The problem statement, all variables and given/known data
Solve 2x''+3x'+40x = 40y+3y'

2. Relevant equations
y = 0.05sin(10t)

3. The attempt at a solution
I used the annihilator method to find the answer of x(t) = Acos(10t)+Bsin(10t)+Ce-0.75tcos(sqrt(311)/4t)+De-0.75tsin(sqrt(311)/4t) where A, B, C and D are constants.

The initial conditions were given as x(0)=0, x'(0)=0, x''(0)=0 and I used 2x''(0)+3x'(0)+40x(0) = 40y(0)+3y'(0) for the last one giving:

0 = A + C

0 = 10B - 0.75C + [sqrt(311)/4]D

0 = 100A +(151/8)C + 3[sqrt(311)/8]D

-160A+30B = 1.5

I know these are right because I double checked with matlab so I'm not really sure what's wrong, but as I said the set is not linearly independent.

2. Apr 16, 2015

### LCKurtz

You mean $Ce^{-\frac 3 4 t}\cos(\frac{\sqrt {311}}{4}t)+De^{-\frac 3 4 t}\sin(\frac{\sqrt {311}}{4}t)$ for the complementary solution $y_c$ part of that.

What isn't linearly independent? The two terms in $y_c$ certainly are.

3. Apr 16, 2015

### vela

Staff Emeritus
The given initial conditions are inconsistent. If you plug in the initial conditions on the lefthand side and set t=0 in the righthand side, you get 0 = 3/2.

The problem is overconstrained as well. If you were to use the method of underdetermined coefficients, you'd only have two undetermined constants, but you have three initial conditions.