# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-5 => Topic started by: Victor Ivrii on November 02, 2018, 03:14:52 PM

Title: Q5 TUT 0201
Post by: Victor Ivrii on November 02, 2018, 03:14:52 PM
Use the method of variation of parameters (without reducing an order) to determine the general solution of the given differential equation:
$$y''' + y' = \tan (t),\qquad -\pi /2 < t < \pi /2.$$
Title: Re: Q5 TUT 0201
Post by: Guanyao Liang on November 02, 2018, 03:54:38 PM
Answer is in the attachment.
Title: Re: Q5 TUT 0201
Post by: 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:

$y = c_1 + c_2\cos(t) + c_3\sin(t) - \ln|\cos(t)| - \sin(t)\ln|\sec(t) + \tan(t)|$
Title: Re: Q5 TUT 0201
Post by: Pengyun Li on November 04, 2018, 01:23:04 AM
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...  :)
Title: Re: Q5 TUT 0201
Post by: Michael Poon on November 04, 2018, 02:31:19 AM
oh right! Totally my bad! time to relearn trig..   :-\