1) 2 d2y/ dt2 + 4 dy/dt – 5y = 0
Reduces
to:
a)
x1' = x2
b)
x2' = -2 x2 + 5 x1/2,
Applying
Laplace transforms to (a):
£{dx1/dt} =
sx1(s) and £ {x2} =
x2(s)
Applying Laplace transforms to (b):
£ {dx2/dt} =
sx2(s)
£ {-2x2
+ 5x1/2} = -2 x2(s)
+ 5x1(s)/ 2
Yielding the two equations:
1) sx1(s) = x2(s)
And
2) sx2(s) = -2
x2(s) + 5x1(s)/ 2
Or:
sx1(s) = x2(s)
5x1(s)/ 2 = x2(s) [s
+ 2]
Applying the Laplace
transform directly to the 2nd order DE:
We have: 2 d2y/ dt2 + 4 dy/dt – 5y = 0
Or: d2y/
dt2 + 2 dy/dt – 5y/2 = 0
£
(d2y/ dt2 ) = s2 y(s) and £ (2 dy/dt ) = 2 s y(s)
With: £(5y/2 )= 5 y(s)/ 2
So rewrite the 2nd order DE in terms of Laplace
transforms only:
s2
y(s) + 2 s y(s) - 5 y(s)/ 2 = 0
Or: ( s2 +
2s – 5/2) y(s) = 0
(2) Propose a way that you'd get to the inverse
Laplace transform, if you approached the 2nd order
problem directly, instead of reducing it to normal form.
We’d
need at least two initial conditions, say y(0) = 2 and y’(0)= 5
Then
we can write: : ( s2 +
2s – 5/2) y(s) = 5s -2
Or: y(s) = (5s -
2)/ ( s2 + 2s – 5/2) =
(5s – 2)/ (s + 2.871) (s – 0.871)
(5s – 2)/ (s + 2.871) (s – 0.871)
For which we’d seek the inverse transform of the last
expression (i.e. from a mathematical tables).
If no direct inverse transform can be found it may be
necessary to reduce the existing transform further say by partial fractions,
i.e.
(5s – 2) = A/ (s +
2.871) + B/ (s – 0.871)
A and B are then found by substituting in the zeros for the
expression (e.g. -2.871 and 0.871) and solving for the A, B then trying to find
the new inverse transform. If none can be found in this case, it may be necessary to employ other methods.
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