A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
solve the differential equation: y''+y=1/sinx
anonymous
 4 years ago
solve the differential equation: y''+y=1/sinx

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0differential equation \(\implies \) lalaly

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5first find the homogeneous solution let\[y''+y=0\]so\[r^2+1=0\]\[r^2=1\]\[r= \pm i\]so\[y_h=c_1cosx+c_2sinx\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5now to find the particular solution we have two methods, which one do u like to use? lol

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5When u decide let me know :D:D im here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks ! i got the homogenous solution but to which methods do you mean ? i tried variation of parameters but got stucked with crazy integral

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5theres method of undetermined coeffecients

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't know how to do it because its 1/sinx and not just sinx, if you can i will be happy if you show me the two ways =]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5ok can u give me sometime i need to do it on my own first then i can write it down here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0great !! thanks alot !!

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5ok variation of parameters first find the wronskian W \[y_1=cosx\]\[y_2=sinx\]dw:1327778897437:dw let \[g(x)=\frac{1}{sinx}\]so the particular solution is\[y_p=y_1 \int\limits{\frac{y_2 \times g(x)}{W}dx}+y_2 \int\limits{\frac{y_1 \times g(x)}{W}dx}\]\[y_p=cosx \int\limits{\frac{sinx \times \frac{1}{sinx}}{1}dx}+sinx \int\limits{\frac{cosx \times \frac{1}{sinx}}{1}dx}\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5\[y_p=cosx \int\limits{dx} +sinx \int\limits{ cotx dx}\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5\[y_p=cosx \times x + sinx \times \log(sinx)\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5\[=xcosx+\log(sinx)sinx\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5so general solution is\[y=y_h+y_p\]\[y=c_1cosx+c_2sinxxcosx+sinx \log(sinx)\]

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.5if u try undetermined coeffecients it gets hairy lol so i prefer this one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks alot !!! highly appreciated =]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.