Use
Laplace transforms to solve the differential equation:
d
3 Y/ dt 3 - d Y /dt =
0
Using the conditions:
Y(0) = 1, Y’(0) =
0 and Y" (0) = 1
Solution: We Write:
£ { d 3 Y/ dt 3 } = s 3 y(s) - Y(0) s 2 - Y'(0) s - Y"(0)
= s 3 y(s) - s 2 - 1
£ {dY / dt} = s y(s) - Y(0) - Y"(0)
= s y(s) - 1
Substitute the expression for each transform:
s 3 y(s) - s 2 - 1 = 0
Collect like terms and transpose:
(s 3 - s) y(s) = s 2 + 1
Solve for y(s):
y(s) = ( s 2 + 1 ) / (s 3 - s) = ( s 2 + 1 ) / s ( s 2 - 1 )
Use of partial fractions yields:
A/ s + B/ ( s 2 - 1 ) = s 2 + 1
Then:
A ( s 2 - 1 ) + Bs = s 2 + 1
Whence:
A s 2 - A + B s = s 2 + 1
Yielding values (by equating coefficients):
A = 1
B - A = 0
So: B = 1
è
( s 2 + 1 ) / s ( s 2 - 1 ) =
1 / s + 1 / ( s 2 - 1 )
We then need to take the inverse Laplace transform :
£ -1 [1 / s + 1 / ( s 2 - 1 ) ]
=
£ -1 [1 / s ] + £ -1 [ 1 / ( s 2 - 1 )]
From a table of transforms (see bottom of previous post from March 2nd):
We see: 1/s is inverse transformed to 1
And similarly: [ 1 / ( s 2 - 1 )] -> cosh (t)
Then the solution to the DE is: Y (t) = 1 + cosh (t)
Check soln. to d 3 Y/ dt 3 - d Y /dt = 0 :
d 3 (1+ cosh (t))/ dt 3 = 0
d (1 + cosh (t))/dt = 0
Then: 0 - 0 = 0
So:
d 3 Y/ dt 3 - d Y /dt = 0
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