Let W_{1} and W_{2} be the subspaces of R^{3} given by
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W_{1} = span |
| and W_{2} = span |
| .
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Find a set of generating vectors for W_{1} \cap W_{2}.
Response. W_{1} and W_{2} are each spanned by two linearly
independent vectors, and for that reason each is a plane through the
origin in R^{3}.
One therefore expects their intersection to be a line through the
origin in R^{3}, i.e., the set of all scalar multiples of a
single vector. The strategy adopted is to find equations defining
W_{1} and W_{2} and then solve both equations simultaneously to
find W_{1} \cap W_{2}.
For W_{1} one seeks a vector (a, b, c) such that
Solving the corresponding linear system of equations using row
operations, one finds that the most general such vector (a, b, c)
is a scalar multiple of (1, -5, 3), and “the” equation of
W_{1} is x - 5 y + 3 z = 0.
A similar method can be used for W_{2}. On the other hand, one
sees quickly by inspection that “the” equation of W_{2} is y = 0.
Then solving the two equations simultaneously -- for example, using
row operations on the matrix
-- one finds that a basis of W_{1} \cap W_{2} is given by the
single vector (-3, 0, 1), i.e.,
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W_{1} \cap W_{2} = span |
| .
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Find a particular solution of the linear differential equation
Response. In the study of differential equations one learns
to look for particular solutions in various ways. In this case it
will be fruitful to propose
as a trial solution. With this y one finds y'' = 2c, and,
therefore,
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y'' + 4 y = (4 a + 2 c) + 4 b x + 4 c x^{2} .
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To have the right-hand side evaluate as x^{2} one needs
Solving this system for a, b, c one obtains
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y = {1}/{4} x^{2} - {1}/{8} .
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