Edwardwen85,
I am going to answer this question from first principles. As a result,
I will include information that one would not normally include if
simply answering a textbook question - rather I will try and explain a
little of what is going on. If this question is for a course, I would
encourage you to reformat the answer to fit the style of the course
(different people teach this material in different ways).
The first step in rewriting this equation is to create a mapping
between the various derivatives of y and the components of the vector
x.
Write x1 = y'', x2 = y', x3 = y.
Then x1' = y''', x2' = y'' = x1, x3' = y' = x2
We can incorporate this information into a matrix as follows :
(x1') (? ? ?) (x1)
(x2') = (1 0 0) (x2)
(x3') (0 1 0) (x3)
All that remains is to represent the actual differential equation,
y''' - 2y'' + 3y' + 4y = 0 in matrix form.
Rearranging and substituting x1', x1, x2, x3 for y''', y'', y' and y
respectively we get :
x1' = 2*x1 - 3*x2 - 4*x3.
Hence the top line of the matrix becomes 2, -3, -4. So the matrix form
of the differential equation is :
(x1') (2 -3 -4) (x1)
(x2') = (1 0 0) (x2)
(x3') (0 1 0) (x3)
This question can be split into two key steps :
1) Write down a relationship between the derivatives of y and
translate them into a simple relationship between elements of the
vector x.
and 2) Rewrite the original differential equation in this form.
(Step 3 is to write out the answer to parts 1 and 2 in matrix form).
Good luck.
calebu2-ga |