0

# Find the general solution?

Find the general solution of x'=(1, sqrt(3), sqrt(3), -1)x+(e^t, sqrt(3)*e^-t).

(this is 2x2 matrix with 1 and sqrt(3) on the left, sqrt(3) and -1 on the right. And 2x1 matrix with e^t on top, sqrt(3)*e^-t on bottom. I know how to find the answer for 2x2 matrix but not the 2x1 matrix. Please help me step by step. The answer for 2x1 matrix is -(2/3, 1/sqrt(3))e^t+(-1, 2/sqrt(3))e^-t where 2/3 is on the top, 1/sqrt(3) is on bottom, and -1 is on top, 2/sqrt(3) is on bottom.)

### 1 Answer by Expert Tutors

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
1
You found the homogeneous solution

xh=c1(-1/sqrt(3),1) e-2t +c2 (sqrt(3),1) e2t

To find the inhomogeneous solution you can use one of two methods
A) undetermined coefficients
B) variation of parameters

I will do method A), perhaps somebody else will do method B) for you.

Your inhomogeneous term is (et, sqrt(3)*e-t), which means x MUST be of the form

x= (A1,A2) et + (B1,B2) e-t

We have to find out what A1, A2, B1, and B2 are. We do this by substituting x and its derivative

x'=(A1,A2) et - (B1,B2) e-t

into the original equation,

x'=(1, sqrt(3), sqrt(3), -1)x+(e^t, sqrt(3)*e^-t)

I will do the matrix term first:

(1, sqrt(3), sqrt(3), -1)x = (1, sqrt(3), sqrt(3), -1) ( (A1,A2) et + (B1,B2) e-t)
= ((A1+sqrt(3)A2)et + (B1+sqrt(3)B2)e-t ,  (sqrt(3) A1 -A2)et + (sqrt(3) B1 -B2)e-t )

Therefore, the first component of x' is
A1 et - B1 e-t = (A1+sqrt(3)A2)et + (B1+sqrt(3)B2)e-t +et

The second component of x' is
A2 et - B2 e-t = (sqrt(3) A1 -A2)et + (sqrt(3) B1 -B2)e-t +sqrt(3)*e-t

Let's collect all terms with an et from the first equation and from the second equation:

A1 = A1+sqrt(3)A2 +1
A2 = sqrt(3) A1 -A2

In the first of these, A1 cancels and we get
A2 = -1/sqrt(3).
Plug this into the second equation and get
A1 = -2/3

Now we need to find B1 and B2. For this we collect all terms with an e-t from the first and the second x' equation:

- B1  = B1+sqrt(3)B2
- B2 = sqrt(3) B1 -B2 +sqrt(3)

In the second of these equations, B2 cancels and we get
B1 = -1
Plug this into the first equation and get
B2 = 2/sqrt(3)

Therefore we found

x= (-2/3,-1/sqrt(3)) et + (-1,2/sqrt(3)) e-t