y'' + 4y = \begin{cases} 3 \sin t, & 0 \le t \le 2\pi \\ 0, & t > 2\pi \end{cases} y(0) = 1, y'(0) = 3
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The given differential equation is 3sin(t). To find the general solution, we need to integrate this equation with respect to t. ∫3sin(t) dt = -3cos(t) + C where C is the constant of integration. Show more…
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