### Author Topic: Q4 TUT 0601  (Read 2504 times)

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2599
• Karma: 0
##### Q4 TUT 0601
« on: October 26, 2018, 05:44:07 PM »
Find the general solution of the given differential equation.
$$y'' + y = \tan (t),\qquad 0< t < \frac{\pi}{2}.$$

#### Jiabei Bi

• Jr. Member
• Posts: 7
• Karma: 10
##### Re: Q4 TUT 0601
« Reply #1 on: October 26, 2018, 08:15:18 PM »
here is my solution

#### Jiacheng Ge

• Full Member
• Posts: 19
• Karma: 8
##### Re: Q4 TUT 0601
« Reply #2 on: October 26, 2018, 09:49:14 PM »
For the homogeneous equation,
r² + 1 = 0
r = i or −i

So, the complementary solution is y = C1cost + C2sint

W[y1,y2] =y1y2' − y1'y2 = cos²t + sin²t = 1

u1(x)=−∫(y2(x)g(x)/W)dx
= −∫(sint tant) dt
= sint − ln(tant+sect)

u2(x)=∫(y1(x)g(x)/W)dx
= ∫(cost tant)dt
= ∫sint dt
= −cos t

So, a particualr solution is
y = u1y1+u2y2
=[sint - ln(tant+sect)]cost − costsint
=−ln(tant+sect)cost

So,the general solution is
y = C1cost + C2sint −ln(tant+sect)cost

« Last Edit: October 26, 2018, 09:51:41 PM by Jiacheng Ge »

#### Zhiya Lou

• Jr. Member
• Posts: 12
• Karma: 12
##### Re: Q4 TUT 0601
« Reply #3 on: October 26, 2018, 09:55:34 PM »
First, solve the homogeneous solution:
$r^2+1=0$
$r_1=i, r_2= -i$
$y_1= cos(t); y_2 = sin(t)$
So, $W=y_1y_2' - y_2y_1'= cos^2(t) + sin^2(t) = 1$

Second, solve for particular solution:
$Y=u_1y_1 + u_2y_2$
$u_1(t) = -\int sin(t)tan(t) dt\ = -\int sec(t)(1-cos^2(t))dt = -\int sec(t) - cos(t) dt = -ln(tan(t) + sec(t)) +sin(t)$

$u_2(t) = \int y_1(t) tan(t) dt = \int cos(t)tan(t) dt = \int sin(t) dt = -cos(t)$

Therefore, $Y= [-ln(tan(t) + sec(t)) +sin(t)](cos(t)) + (-cos(t)) (sin(t))$

General Solution:
$y(t)= c_1cos(t) + c_2sin(t) - ln(tan(t)+sec(t))(cos(t))$

#### Victor Ivrii

• Administrator
• Elder Member
• Posts: 2599
• Karma: 0
##### Re: Q4 TUT 0601
« Reply #4 on: October 27, 2018, 12:38:58 PM »
Jiacheng, unreadable

Zhiya
Need to type \sin x, \ln x ...