Math  /  Calculus

QuestionTransform the differential equation 3y+2y+4y=emty(0)=2y=4\begin{array}{c} -3 y^{\prime \prime}+2 y^{\prime}+4 y=e^{m t} \\ y(0)=2 \\ y^{\prime}=-4 \end{array} into an algebraic equation by taking the Laplace transform of each side. \square == \square Therefore Y=Y= \square

Studdy Solution
Therefore, Y=6s2+(6m8)s+8m+1(sm)(3s2+2s+4)Y = \frac{-6s^2 + (6m - 8)s + 8m + 1}{(s-m)(-3s^2 + 2s + 4)}

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