MAT244--2018F > Quiz-5

Q5 TUT 0201

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**Victor Ivrii**:

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.

$$

**Guanyao Liang**:

Answer is in the attachment.

**Michael Poon**:

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:

$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$

**Pengyun Li**:

--- Quote from: Michael Poon on November 02, 2018, 04:25:48 PM ---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:

$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$

--- End quote ---

Hi Michael, $\ln{|sec(t)|} and -\ln{|cos(t)|}$ are the same thing... :)

**Michael Poon**:

oh right! Totally my bad! time to relearn trig.. :-\

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