Find the solution of the initial value problem $y'' + 2y' + 5y = 16e^{-t}\cos(2t)$, $y(0) = 4$, $y'(0) = 0$. $y(t) = $
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The roots are $r = \frac{-2 \pm \sqrt{4 - 4(5)}}{2} = -1 \pm 2i$. Thus, $y_c(t) = e^{-t}(c_1 \cos(2t) + c_2 \sin(2t))$. Show more…
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Find the solution of the initial value problem y'' + 2y' + 5y = 16e^(-t)cos(2t), y(0) = 6, y'(0) = 0
Madhur L.
Find the solution of the initial value problem y'' + 2y' + 5y = 16e^{-t} cos (2t), y (0) = 6, y' (0) = 0. Enclose arguments of functions in parentheses. For example, sin (2x).
Supreeta N.
$$\begin{array}{l}{y^{\prime \prime \prime}+y^{\prime \prime}+3 y^{\prime}-5 y=16 e^{-t}} \\ {y(0)=0, \quad y^{\prime}(0)=2, \quad y^{\prime \prime}(0)=-4}\end{array}$$
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