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Consider the initial value problem y′′+36y=2cos(6t),y(0)=0,y′(0)=0. y″+36y=2cos⁡(6t),y(0)=0,y′(0)=0. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) b

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LammettHash

Recall the Laplace transform of a second-order derivative,

[tex]L(y''(t)) = s^2Y(s)-sy(0)-y'(0)[/tex]

and the transform of cosine,

[tex]L(\cos(at))=\dfrac s{a^2+s^2}[/tex]

Here, both [tex]y(0)=y'(0)=0[/tex], so taking the transform of both sides of

[tex]y''(t)+36y(t)=2\cos(6t)[/tex]

gives

[tex]s^2Y(s)+36Y(s)=\dfrac{2s}{36+s^2}[/tex]

[tex]\implies Y(s)=\dfrac{2s}{(s^2+36)^2}[/tex]

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