Solve the initial value problem: dy = 3e^(3t) sin(e^(3t) - 2), y(h32) = 0. y(t)
Added by Julia J.
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Let u = e^(3t) - 2, then du/dt = 3e^(3t), and we can rewrite the differential equation as: dy/du * du/dt = 3e^(3t)sin(u) dy/du = sin(u) Integrating both sides with respect to u, we get: y = -cos(u) + C where C is a constant of integration. Show more…
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