Resulting system of equations is not linearly independent

In summary, the conversation discusses solving a differential equation using the annihilator method and finding the complementary solution. The given initial conditions are inconsistent and the problem is overconstrained, leading to difficulties in finding a solution.
  • #1
e^(i Pi)+1=0
247
1

Homework Statement


Solve 2x''+3x'+40x = 40y+3y'

Homework Equations


y = 0.05sin(10t)

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.
 
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  • #2
e^(i Pi)+1=0 said:

Homework Statement


Solve 2x''+3x'+40x = 40y+3y'

Homework Equations


y = 0.05sin(10t)

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.

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.

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.

What isn't linearly independent? The two terms in ##y_c## certainly are.
 
  • #3
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.
 

What does it mean for a resulting system of equations to not be linearly independent?

When a system of equations is not linearly independent, it means that one or more of the equations in the system can be derived or expressed as a combination of the other equations. This makes the system redundant and reduces the number of unique equations in the system.

What are the implications of having a resulting system of equations that is not linearly independent?

When the resulting system of equations is not linearly independent, it means that there is not enough information to solve the system and find a unique solution. The system may have infinitely many solutions or no solutions at all.

How can I determine if a resulting system of equations is linearly independent?

To determine if a resulting system of equations is linearly independent, you can use the row reduction method to put the system in row-echelon form. If any row is entirely zeros or if a row is a multiple of another row, then the system is not linearly independent.

What is the importance of having a linearly independent system of equations?

A linearly independent system of equations is important because it allows us to find a unique solution to the system. This is crucial in many applications of mathematics, such as in engineering, physics, and economics, where finding a single solution is necessary.

What can I do if I have a resulting system of equations that is not linearly independent?

If your resulting system of equations is not linearly independent, you can use additional equations or information to make the system linearly independent. This can be done by either adding new equations or eliminating redundant equations from the system.

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