InterviewSolution
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Solve the differential equation d2y/dt2 + 6dy/dt + 8y = 5sin3t using classical methods.\(\frac{d^2y}{dt^2} + \frac{6dy}{dt} + 8y = 5 sin 3t\) |
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Answer» Given differential equation is \(\frac{d^2y}{dt}^2 + 6 \frac{dy}{dt} + 8y = 5 sin\,3t\) \(\Rightarrow\) D2y + 6Dy + 8y = 5 sin 3t (\(\because \frac{d}{dt} = D\) & \(\frac{d^2}{dt^2} = D^2\)) \(\Rightarrow\) (D2 + 6D + 8)y = 5sin 3t The characterstic equation of given differential equation is \(\Rightarrow\) m2 + 6m + 8 \(\Rightarrow\) m2 + 2m + 4m + 8 = 0 \(\Rightarrow\) m(m + 2) + 4(m + 4) = 0 \(\Rightarrow\) (m + 2)(m + 4) = 0 \(\Rightarrow\) m = -2, -4 The complementary function is C. F = C1e-2t + C2e-4t Now, particular integral is P.I = \(\frac{1}{D^2 + 6D + 8} \) 5 sin 3t = \(\frac{1}{-9 + 6D + 8}\) 5 sin 3t (\(\frac{1}{f(D)^2} = \frac{1}{f(-a)^2} sin ax; \, f(-a^2) \neq 0\)) = \(\frac{6D + 1}{(6D - 1)(6D + 1)}\) 5sin 3t = \(\frac{6D + 1}{36D^2 - 1}\) 5 sin 3t = 5 (6D + 1) \(\frac{sin \,3t}{-36 \times 9-1}\) = \(\frac{-5}{325}\) (6D + 1) sin 3t = \(-\frac{1}{65} (18 cos\, 3t + sin \, 3t)\) Hence the solution is y = C.F + P.I \(\Rightarrow\) y = C1e-2t + C2e-4t - \(\frac{1}{65}\)(18 cos 3t + sin 3t) |
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