How to Solve a System of Equations with Multiple Variables and Equations

In summary, the person is trying to solve a differential equation, but they are not sure how to do it. They are stuck on a system of equations, but they are not sure how to solve for the coefficients. They attempted to solve it using a calculator, but it gave them an error. They needed to allow one of the variables to be arbitrary in order to solve for the rest of the coefficients.
  • #1
ns5032
28
0

Homework Statement



Well, I am in the middle of solving a (quite difficult) differential equation using the Method of Undetermined Coefficients, and have finally come across a system of equations and I need to solve for A(sub 1), or (A1), (A2), (B1), and (B2), but I cannot figure out how to (or if it is possible at all, seeing as how I could have done the rest of the problem wrong before it). The system of equations I now have is:

(1) -7(A1) + 3(A2) + 6(B1) - 21(B2) = 2
(2) 3(A1) + 12(A2) - 7(B1) + 3(B2) = -34
(3) 21(A1) + 5(B1) = 150

Homework Equations


---

The Attempt at a Solution



Using equation (3), I can determine that
(A1) = (50/7) - (5/21)(B1)
and that
(B1) = 30 - (21/5)(A1)

Using these and plugging them into equation (2), I get a 4th equation:

(4) 1029(A1) + 420(A2) - 25(B1) + 105(B2) = 5410

Now I have a new system of equations:

(1) -7(A1) + 3(A2) + 6(B1) - 21(B2) = 2
(2) 3(A1) + 12(A2) - 7(B1) + 3(B2) = -34
(3) 21(A1) + 5(B1) = 150
(4) 1029(A1) + 420(A2) - 25(B1) + 105(B2) = 5410

Now, I tried using my calculator to solve this system, but it only gives me an error. Is there another way to solve for the coefficients?? Any help would be great!
 
Physics news on Phys.org
  • #2
Hi,
I'm not the best at linear algebra, but it looks to me like you have 3 equations and 4 unknowns. That can't be solved. You do need one more equation, but it seems to me like if you solve the last equation for A1 and A2 and then plug that back into equation 2, you have a dependent system. If you solve number 3 for A1 and then plug that into 1 and 2, then you have 2 equations and 3 unknowns...still can't solve it.

I may be wrong here...I'm wrong a lot,
But that's what I remember.
CC
 
  • #3
Well I started with only 3 equations, but under "attempt at the solution" I did what you said and got a 4th equation. So now I have 4 equations with 4 variables which should work, but when I use matrices in my calculator, it won't solve it.
 
  • #4
You have to allow one of the variables to be arbitrary, and then solve for the rest of them in terms of the arbitrary variable. This will give you a set of solutions, which are known as the "solution space"
 
  • #5
ns5032 said:
Well I started with only 3 equations, but under "attempt at the solution" I did what you said and got a 4th equation. So now I have 4 equations with 4 variables which should work, but when I use matrices in my calculator, it won't solve it.

What you did to obtain that 4th equation was putting A1 and B1 from (3) into another equation. While that may appear to give you a fourth equation, if you actually try to solve it via row-reduction, you should see that you will get a row of zero's. Meaning that you will have an infinite number of solutions

ns5032 said:
but I cannot figure out how to (or if it is possible at all, seeing as how I could have done the rest of the problem wrong before it)
You could post the differential equation you tried to solve and then post your attempt at it and we can attempt to see if you went wrong somewhere in the process.
 
Last edited:

1. What is a system of equations?

A system of equations refers to a set of multiple equations that are analyzed together to determine the values of the variables that satisfy all of the equations simultaneously.

2. How can I solve a system of equations?

There are various methods for solving a system of equations, including substitution, elimination, and graphing. Each method involves manipulating the equations to isolate a variable and then using that value to solve for the other variables.

3. Can a system of equations have more than one solution?

Yes, a system of equations can have one, infinite, or no solutions. One solution means that there is a unique set of values that satisfy all of the equations, infinite solutions means that all values of the variables satisfy the equations, and no solution means that there are no values that satisfy the equations.

4. What is the importance of solving systems of equations?

Solving systems of equations is important in many fields of science and mathematics, as it allows for the determination of multiple unknown variables and can be used to model and predict real-world situations. It is also a fundamental concept in linear algebra and is used extensively in engineering and economics.

5. Are there any tips for solving systems of equations more efficiently?

Yes, there are a few tips that can help make solving systems of equations more efficient. One tip is to start by isolating a variable with a coefficient of 1, if possible. Also, try to eliminate one variable at a time to make the process less overwhelming. Lastly, double-check your solutions by plugging them back into the original equations to ensure they satisfy all of the equations.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
949
  • Calculus and Beyond Homework Help
Replies
1
Views
987
Replies
7
Views
578
  • Calculus and Beyond Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
583
  • Quantum Physics
Replies
32
Views
2K
  • Quantum Interpretations and Foundations
2
Replies
38
Views
4K
  • Calculus and Beyond Homework Help
Replies
9
Views
698
  • Calculus and Beyond Homework Help
Replies
11
Views
3K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
Back
Top