Toronto Math Forum
MAT244--2018F => MAT244--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 02, 2018, 03:14:52 PM
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Use the method of variation of parameters (without reducing an order) to determine the general solution of the given differential equation:
$$
y''' + y' = \tan (t),\qquad -\pi /2 < t < \pi /2.
$$
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Answer is in the attachment.
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Hi Guanyao Liang,
Your answer is very close, but I think you messed up a calculation. The integral of $\tan(t)$ is $-\ln|\cos(t)|$, not $\ln|\sec(t)|$.
Therefore the solution should be:
$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$
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Hi Guanyao Liang,
Your answer is very close, but I think you messed up a calculation. The integral of $\tan(t)$ is $-\ln|\cos(t)|$, not $\ln|\sec(t)|$.
Therefore the solution should be:
$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$
Hi Michael, $\ln{|sec(t)|} and -\ln{|cos(t)|}$ are the same thing... :)
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oh right! Totally my bad! time to relearn trig.. :-\