Please Find the particular solution and explain the answer step-by-step: y\'\'\'

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Let's assume the following initial values: y'''(0) = y''(0) = y'(0) = y(0) = 0. (shouldn't they be included in this assignment?) Transform with Laplace, y(t) --> Y(s) and cos(t) + cosh(2t) --> X(s): [ s^4 Y(s) - s³ y(0) - s² y'(0) - s y''(0) - y'''(0) ] - 3 [ s² Y(s) - s y(0) - y'(0) ] - 4 Y(s)= X(s) s^4 Y(s) - 3 s² Y(s) - 4 Y(s) = X(s) (s^4 - 3 s² - 4) Y(s) = X(s) Look up X(s) in table: X(s) = s / (s² + 1) + s / (s² - 4), so Y(s) = [ s / (s² + 1) + s / (s² - 4) ] / (s^4 - 3 s² - 4) Try to find the reverse transformation of that function. First simplify: Y(s) = s (2s² - 3) / ( (s² + 1) (s² - 4) (s^4 - 3 s² - 4) ) Expand to partial fractions, for this we need to factor that last expression on the right: s^4 - 3 s² - 4 = 0 --> +/- 2, +/- i --> s^4 - 3 s² - 4 = (s - 2) (s + 2) (s² + 1) Y(s) = s (2s² - 3) / ( (s² + 1) (s² - 4) (s - 2) (s + 2) (s² + 1) ) = 1/40 [ 1/(s-2)² - 1/(s+2)² ] - 1/5 s/(s²+1)² Inverse Laplace transformation y(t) = t [sinh(2 t)/2 - sin(t) ] / 10 --> y solves both y''''-3y''-4y = cos(t) + cosh(2t) and y'''(0) = y''(0) = y'(0) = y(0) = 0.

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