Q5 TUT 0201

[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]

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... 

[Michael Poon]

oh right! Totally my bad! time to relearn trig..