Question

Use the laplace transform to solve the given initial-value problem. y'' − 6y' 13y = 0, y(0) = 0, y'(0) = −5

Answer 1

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

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Answer 2

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:

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