Find the general solution?

Question

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.)

 

Answer 1

You found the homogeneous solution

 

x_h=c₁(-1/sqrt(3),1) e^-2t +c₂ (sqrt(3),1) e^2t

 

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 (e^t, sqrt(3)*e^-t), which means x MUST be of the form

 

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

= ((A1+sqrt(3)A2)e^t + (B1+sqrt(3)B2)e^-t ,  (sqrt(3) A1 -A2)e^t + (sqrt(3) B1 -B2)e^-t )

 

Therefore, the first component of x' is

A1 e^t - B1 e^-t = (A1+sqrt(3)A2)e^t + (B1+sqrt(3)B2)e^-t +e^t

 

The second component of x' is

A2 e^t - 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 e^t 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)) e^t + (-1,2/sqrt(3)) e^-t

 

which is your solution!