## Saturday, August 3, 2013

### Solutions to Rewriting DEs as Laplace Transforms

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)

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.