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anonymous

  • 4 years ago

solve the differential equation: y''+y=1/sinx

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  1. anonymous
    • 4 years ago
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    differential equation \(\implies \) lalaly

  2. lalaly
    • 4 years ago
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    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
    • 4 years ago
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    now to find the particular solution we have two methods, which one do u like to use? lol

  4. lalaly
    • 4 years ago
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    When u decide let me know :D:D im here

  5. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    theres method of undetermined coeffecients

  7. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    ok can u give me sometime i need to do it on my own first then i can write it down here

  9. anonymous
    • 4 years ago
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    great !! thanks alot !!

  10. lalaly
    • 4 years ago
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    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
    • 4 years ago
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    \[y_p=-cosx \int\limits{dx} +sinx \int\limits{ cotx dx}\]

  12. lalaly
    • 4 years ago
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    \[y_p=-cosx \times x + sinx \times \log(sinx)\]

  13. lalaly
    • 4 years ago
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    \[=-xcosx+\log(sinx)sinx\]

  14. lalaly
    • 4 years ago
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    so general solution is\[y=y_h+y_p\]\[y=c_1cosx+c_2sinx-xcosx+sinx \log(sinx)\]

  15. lalaly
    • 4 years ago
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    if u try undetermined coeffecients it gets hairy lol so i prefer this one

  16. anonymous
    • 4 years ago
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    thanks a-lot !!! highly appreciated =]

  17. lalaly
    • 4 years ago
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    your welcome :D:D

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