### Author Topic: Q5 TUT 0201  (Read 2054 times)

#### Victor Ivrii ##### Q5 TUT 0201
« 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:
$$y''' + y' = \tan (t),\qquad -\pi /2 < t < \pi /2.$$

#### Guanyao Liang

• Jr. Member
•  • Posts: 13
• Karma: 12 ##### Re: Q5 TUT 0201
« Reply #1 on: November 02, 2018, 03:54:38 PM »

#### Michael Poon ##### Re: Q5 TUT 0201
« Reply #2 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:

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

#### Pengyun Li

• Full Member
•   • Posts: 20
• Karma: 14 ##### Re: Q5 TUT 0201
« Reply #3 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... #### Michael Poon ##### Re: Q5 TUT 0201
« Reply #4 on: November 04, 2018, 02:31:19 AM »
oh right! Totally my bad! time to relearn trig.. 