How many equations are needed to solve this system?

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In summary, the conversation discusses a system of linear differential equations that can be solved using the method of undetermined coefficients. The system has four equations and four unknowns, and the solution is a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3. Various approaches to solving the system are suggested, including substitution and checking the determinant of the matrix of coefficients.
  • #1
enceladus_
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This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.
 
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  • #2
You can solve your third equation directly:

-a1 + 5b1 +a1 + 1 = 0 → 5b1 + 1 = 0 → b1 = -1/5

This contradicts the answer for b1 you stated. There is either a mistake in the system or in the solution.

Junaid Mansuri
 
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  • #3
It should be:

-a1 +5b1 + a2 + 1 = 0

My mistake.
 
  • #4
An easy approach to solve this system is through substitution. Solve for a1 in equation 4 (in terms of the b's). Solve for a2 in equation 2 (in terms of the b's). Substitute the expressions for a1 and a2 into equations 1 and 3 and you will then have a system of two equations with only b1 and b2 in them which is easy to solve. Then you can use the known values of b1 and b2 to directly find a1 and a2 from the expressions you had before. I think this method would be easier than elimination.
 
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  • #5
enceladus_ said:
This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.

It looks like you have four equations in four unknowns (a1, a2, b1, and b2). Some of the variables just have 0 coefficients in each of the equations. As long as the determinant of the matrix of coefficients for the system is non-zero, you can obtain a unique solution for the system.
 
  • #6
Many thanks, I was able to solve it!
 
  • #7
enceladus_ said:
This system originates from systems of linear Diff Eqs. I am required to use the method of undetermined coefficients.

-a2 + 5b2 - a1 = 0
-a2 + b2 -b1 -2 = 0
-a1 + 5b1 +a2 + 1 = 0
-a1 + b1 + b2 = 0

I can't figure out how I am able to solve this. In each case, I have two equations and three unknowns. I know the system is correct.

The answers are:

a2 = -3, b2 = -2/3, a1 = -1/3, b1 = 1/3.

Thanks in advance.

I see four equations in four unknowns?
 
  • #8
mathman said:
I see four equations in four unknowns?

You're late to the party. See post #5 above.
 

What is a system of equations?

A system of equations is a set of two or more equations with multiple variables that need to be solved simultaneously. The solution to a system of equations is a set of values that satisfy all of the equations in the system.

How do you solve a system of equations?

There are several methods for solving a system of equations, such as substitution, elimination, and graphing. The most common method is substitution, where one equation is solved for one variable and then substituted into the other equations to solve for the remaining variables.

What is the importance of solving systems of equations?

Solving systems of equations is essential in many fields, such as science, engineering, economics, and statistics. It allows us to find the relationships between variables and make predictions or solve real-world problems.

Can systems of equations have more than two variables?

Yes, systems of equations can have any number of variables. However, the number of equations in the system must be equal to the number of variables for a unique solution to exist.

Are there any online tools available for solving systems of equations?

Yes, there are many online calculators and tools available for solving systems of equations. These tools use advanced algorithms and techniques to quickly and accurately solve systems of equations for any number of variables.

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