1) Solve:
d 3 Y/ dt 3 - d Y /dt = 0
Using the Laplace transform and 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 the table of transforms (Feb. 7 post) :
We see: 1/s is inverse transformed to 1
And similarly: [ 1 / ( s 2 - 1 )] -> cosh (t)
Then the solution to the DE is:
£ { 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 the table of transforms (Feb. 7 post) :
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)
2) Solve:
d2 y/ dt 2 + 4y = 3 sin t
Using the appropriate Laplace transform:
With y = F(t), F(0) = 1, F'(t) = 0
Solution:
Write:
4 £ {F (t) } = 4 F(s)
Then:
Using partial fractions we get:
F(s) = 3 / (s 2 + 1) (s 2 + 4) + s /(s 2 + 4)
= 1 / (s 2 + 1) - 1/ (s 2 + 4) + s /(s 2 + 4)
After taking inverse transforms:
F(t) =
sin t - ½ sin
2t + cos 2t
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