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Answer:
Laplace transforms turn a Differential equation into an algebraic, so we can solve easier.
y'= pY-y(0)
y"=p²Y - py(0)- y'(0)
Substituting these in differential equation :
p²Y -py (0) -y' (0)-6(pY-y(0)) + 13Y
Substituting in the initial conditions given , fact out Y, and get:
Y( p²-6p+13) = -3
Y=-3/ p²-6p+13
now looking this up in a table to Laplace transformation we get:
y=-3/2.e³т sin(2t)
for the last one, find the Laplace of t∧2 . which is 2/p³
pY - y(0)+ 5Y= 2/p³
Y= 2/p³(p+5)
Taking partial fractions:
Y=-2/125(p+5) + 2/125p - 2/25p² + 2/5p³
Learn more about differential equation here:
https://brainly.com/question/14620493
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Answer:
The integral transform that converts a function of a real variable to a function of a complex variable is called Laplace transform. first we need to substitute y' and y" in differential equation then finding Laplace transformation and at last taking partial fractions.
Given: y'= pY-y(0)
y"=p²Y - py(0)- y'(0)
Putting y' and y" in differential equation :
p²Y -py (0) -y' (0)-6(pY-y(0)) + 13Y
Substituting in the initial conditions given , fact out Y, and get:
Y( p²-6p+13) = -3
Y=-3/ p²-6p+13
by Laplace transform we get:
y=-3/2.e³т sin(2t)
for the last one, find the Laplace transform of t∧2 . which is 2/p³
pY - y(0)+ 5Y= 2/p³
Y= 2/p³(p+5)
Taking partial fractions:
Y=-2/125(p+5) + 2/125p - 2/25p² + 2/5p³
Learn more about Laplace transform here:
https://brainly.com/question/1597221
#SPJ4
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