1) 2 d

^{2}y/ dt^{2}+ 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 2

^{nd}order DE:
We have: 2 d

^{2}y/ dt^{2}+ 4 dy/dt – 5y = 0
Or: d

^{2}y/ dt^{2}+ 2 dy/dt – 5y/2 = 0
£
(d

^{2}y/ dt^{2 }) = s^{2}y(s) and £ (2 dy/dt ) = 2 s y(s)
With: £(5y/2 )= 5 y(s)/ 2

So rewrite the 2

^{nd}order DE in terms of Laplace transforms only:
s

^{2}y(s) + 2 s y(s) - 5 y(s)/ 2 = 0
Or: ( s

^{2}+ 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: : ( s

^{2}+ 2s – 5/2) y(s) = 5s -2
Or: y(s) = (5s -
2)/ ( s

(5s – 2)/ (s + 2.871) (s – 0.871)

^{2}+ 2s – 5/2) =(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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