## anonymous 4 years ago solve the differential equation: y''+y=1/sinx

1. anonymous

differential equation $$\implies$$ lalaly

2. lalaly

first 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$

3. lalaly

now to find the particular solution we have two methods, which one do u like to use? lol

4. lalaly

When u decide let me know :D:D im here

5. anonymous

thanks ! i got the homogenous solution but to which methods do you mean ? i tried variation of parameters but got stucked with crazy integral

6. lalaly

theres method of undetermined coeffecients

7. anonymous

i 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 =]

8. lalaly

ok can u give me sometime i need to do it on my own first then i can write it down here

9. anonymous

great !! thanks alot !!

10. lalaly

ok 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}$

11. lalaly

$y_p=-cosx \int\limits{dx} +sinx \int\limits{ cotx dx}$

12. lalaly

$y_p=-cosx \times x + sinx \times \log(sinx)$

13. lalaly

$=-xcosx+\log(sinx)sinx$

14. lalaly

so general solution is$y=y_h+y_p$$y=c_1cosx+c_2sinx-xcosx+sinx \log(sinx)$

15. lalaly

if u try undetermined coeffecients it gets hairy lol so i prefer this one

16. anonymous

thanks a-lot !!! highly appreciated =]

17. lalaly