anonymous
  • anonymous
solve the differential equation: y''+y=1/sinx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
differential equation \(\implies \) lalaly
lalaly
  • lalaly
first find the homogeneous solution let\[y''+y=0\]so\[r^2+1=0\]\[r^2=-1\]\[r= \pm i\]so\[y_h=c_1cosx+c_2sinx\]
lalaly
  • lalaly
now to find the particular solution we have two methods, which one do u like to use? lol

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lalaly
  • lalaly
When u decide let me know :D:D im here
anonymous
  • 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
lalaly
  • lalaly
theres method of undetermined coeffecients
anonymous
  • 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 =]
lalaly
  • lalaly
ok can u give me sometime i need to do it on my own first then i can write it down here
anonymous
  • anonymous
great !! thanks alot !!
lalaly
  • 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\[y_p=-y_1 \int\limits{\frac{y_2 \times g(x)}{W}dx}+y_2 \int\limits{\frac{y_1 \times g(x)}{W}dx}\]\[y_p=-cosx \int\limits{\frac{sinx \times \frac{1}{sinx}}{1}dx}+sinx \int\limits{\frac{cosx \times \frac{1}{sinx}}{1}dx}\]
lalaly
  • lalaly
\[y_p=-cosx \int\limits{dx} +sinx \int\limits{ cotx dx}\]
lalaly
  • lalaly
\[y_p=-cosx \times x + sinx \times \log(sinx)\]
lalaly
  • lalaly
\[=-xcosx+\log(sinx)sinx\]
lalaly
  • lalaly
so general solution is\[y=y_h+y_p\]\[y=c_1cosx+c_2sinx-xcosx+sinx \log(sinx)\]
lalaly
  • lalaly
if u try undetermined coeffecients it gets hairy lol so i prefer this one
anonymous
  • anonymous
thanks a-lot !!! highly appreciated =]
lalaly
  • lalaly
your welcome :D:D

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