Thread: show vector satisfies given DE

1. show vector satisfies given DE

Hi,

I am very rusty with my math, and need help understanding how to approach this type of problem.

(problem)
verify that the given vector satisfies the given differential equation:

x'=Ax
where A is a 3x3 matrix

A = {1,1,1;2,1,-1;0,-1,1}

and x = Be^(-t) + 2Ce^(2t)
where B and C are 3x1 matrices

B = {6;-8;4}
C = {0;1;-1}
(\problem)

I differentiated the x with respect to t, but I dont know how to show that the given vector x' fits the DE.

Thanks,
OkashiiKen

2. nm, solved the problem

feeling rusty.....

x' = Ax, so A^(-1)x' = x

from there I can show the given vector satisfies the DE provided.

3. Originally Posted by OkashiiKen
Hi,

I am very rusty with my math, and need help understanding how to approach this type of problem.

(problem)
verify that the given vector satisfies the given differential equation:

x'=Ax
where A is a 3x3 matrix

A = {1,1,1;2,1,-1;0,-1,1}

and x = Be^(-t) + 2Ce^(2t)
where B and C are 3x1 matrices

B = {6;-8;4}
C = {0;1;-1}
(\problem)

I differentiated the x with respect to t, but I dont know how to show that the given vector x' fits the DE.

Thanks,
OkashiiKen
We have a proposed solution x:
$x = \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}$

So
$x' = - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}$

Thus we need to verify the equation:
$x' = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) x$

$- \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} =$ $\left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )$

The RHS becomes:
$\left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} \right )$ $+ \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )$

$= \left ( \begin{array}{c} 2 \\ 0 \\ 12 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 2 \\ -2 \end{array} \right ) e^{2t}$

We can factor a 2 from each of the 3 x 1 matrices, giving:
$= 2 \left ( \begin{array}{c} 1 \\ 0 \\ 6 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}$

which is not x'. This shows that the given x is NOT a solution to the differential equation.

-Dan

4. We have a proposed solution x:
x = \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

So
x' = - \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

Thus we need to verify the equation:
x' = \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) x

- \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} =\left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

The RHS becomes:
\left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( \left ( \begin{array}{c} 6 \\ -8 \\ 4 \end{array} \right ) e^{-t} \right )+ \left ( \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 0 & -1 & 1 \end{array} \right ) \left ( 2 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t} \right )

= \left ( \begin{array}{c} 2 \\ 0 \\ 12 \end{array} \right ) e^{-t} + 2 \left ( \begin{array}{c} 0 \\ 2 \\ -2 \end{array} \right ) e^{2t}

We can factor a 2 from each of the 3 x 1 matrices, giving:
= 2 \left ( \begin{array}{c} 1 \\ 0 \\ 6 \end{array} \right ) e^{-t} + 4 \left ( \begin{array}{c} 0 \\ 1 \\ -1 \end{array} \right ) e^{2t}

which is not x'. This shows that the given x is NOT a solution to the differential equation.

-Dan
Thanks for your solution, and you are correct. I mistyped the problem 3x1 matrix so that the given vector does not satisfy the give DE. Vector B should be {6;-8;-4} and NOT {6;-8;4}. Using the correct vector, and following your methodology:

A = {1,1,1;2,1,-1;0,-1,1}

x' = Ax
= A{6;-8;-4}e^(-t) + 2{0;1;-1}e^(2t)
= {-6;8;4}e^(-t) + 2{0;2;-2}e^(2t)
= {-6;8;4}e^(-t) + {0;4;-4}e^(2t)

furthermore, by deriving x to get x' we see that
x' = -{6;-8;-4}e^(-t) + 2*2{0;1;-1}e^(2t)
= {-6;8;4}e^(-t) + {0;4;-4}e^(2t)

I initially did this using A^-1, which was not needed, but still worked. Wasnt really sure what I was doing. Thanks for your help!