Use the Laplace transform to solve the following initial value problem: y"-4y'+5y=0 y(0)=1, y'(0)=2
Using Y for the Laplace transform of y(t), i. e. Y=L{y(t)},
find the equation you get by taking the Laplace transform of the differential equation
=0
Now solve for Y(s)=
By completing the square in the denominator and inverting the transform, find y(t)=

Laplace transform of the original differential equation:

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Given that and , solve for to obtain:

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Apply inverse Laplace transform to obtain:

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Step-by-step explanation:

Apply these two rules to replace all , , and in the original equation with their Laplace transforms:

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Substitute in the values and .

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Solve for after rearranging this equation:

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Note that if denominator is the left-hand side of a quadratic equation, this equation would have no real root. Hence, complete the square in the denominator:

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Look up a table of Laplace transforms. Apply the rule , where and .

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$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

answer: 8d

step-by-step explanation:

45+5= 50+5=55+5=60