# How to solve this system

1. Aug 5, 2013

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.

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

Last edited: Aug 5, 2013
2. Aug 5, 2013

### junaid314159

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

Last edited: Aug 5, 2013
3. Aug 5, 2013

It should be:

-a1 +5b1 + a2 + 1 = 0

My mistake.

4. Aug 5, 2013

### junaid314159

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.

Last edited: Aug 5, 2013
5. Aug 5, 2013

### SteamKing

Staff Emeritus
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. Aug 6, 2013

Many thanks, I was able to solve it!

7. Aug 6, 2013

### mathman

I see four equations in four unknowns???

8. Aug 6, 2013

### SteamKing

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