y"+y=csc^3xcotx VARIATION OF PARAMETERS :)) i have an answer but the problem is the process in getting the answer. the wolframalpha has an incorrect answer :D @TuringTest :D
\[y_c=c_1\cos x+c_2\cos x\]got the wronskian yet?
y=C1cosx+C2sinx+1/6cotxcscx that's my answer but i dont know where does 1/6cotxcscx came from. :)
whoa one of those should be sin
yea the wronkian is 1
let me see...
i got the answer from the book but my book has no solution. tsk
\[y_p-\cos x\int \csc^2x\cot xdx+\sin x\int\csc^4 x\cos^2 xdx\]unless I made a mistake...
yp=
first integral is easy, the next will clearly require some trig manipulation
yea. i find hard time in the next integration. can you help me with this? :)
I'm trying to think of the smart way to do it...
HAHA. whoa. i hope your smart brain can help with this. :)
across could do this in her sleep, yet she remains silent....
HAHAHA. she's just viewing my problem :(( urgghh.
oh we can rewrite this as\[\int\csc^2x\cot^2xdx\]as well, so it's doable by u-sub
\[u=\cot x\implies du=-\csc^2x dx\]
\[-\int u^2du=-\frac13\cot^3x+C\]
you can do the other integral, right?
nope. HAHA. can you finish it? :)
try the exact same u-sub!\[\int\csc^2x\cot x dx\]\[u=\cot x\implies du=-\csc^2x\]
-1/2cot^2x+C am i right?
yep
so now plug all that stuff into the formula
so the yp= -1/3cot^3x-1/2cot^2x? and the answer will be ???
you forgot to multiply by y1 and y2 repectively
yp= -1/3sinx*cot^3x+1/2cosx*cot^2x looks like they did some fancy trig identities from there maybe
I gotta eat breakfast, good luck simplifying!
okay. thanks. :) happy eating
the answer will be yp= 1/6 cotxcscx????
hopefully
awwwweee.??? so my solution will be wronskian, integration then yp and lastly y? right?
yep 1)find complimentary solution yc 2)find wronskian 3)apply formula for variation of parameters to get the particular solution yp 4)add the complimentary and particular to find y=yc+yp
can you solve my another problem that i linked you? :) the reduction of order i have the same problem with this. i mean im figuring out the process in getting the same answer :)
I'm pretty much out the door to class... you can post and maybe I'll help, but I'm not the only one here who can do DE's
opppss... sure2. thanks for the info :)
welcome :) just post separately, I bet across will help
To avoid redundancy, I will just remind you that you are looking for a particular solution\[y_p(x)=-y_1(x)\int\frac{y_2(x)g(x)}{W(y_1,y_2)(x)}dx+y_2(x)\int\frac{y_1(x)g(x)}{W(y_1,y_2)(x)}dx.\]In this case,\[y_1(x)=\cos(x),\]\[y_2(x)=\sin(x),\]\[g(x)=\csc^3(x)\cot(x)\text{, and}\]\[W(y_1,y_2)=\begin{vmatrix}\cos(x)&\sin(x)\\-\sin(x)&\cos(x)\end{vmatrix}=\cos^2(x)+\sin^2(x)=1.\]Therefore,\[y_p(x)=\sin(x)\int\cos(x)\csc^3(x)\cot(x)dx-\cos(x)\int\sin(x)\csc^3(x)\cot(x)\]\[=\frac{1}{2}\cos(x)\cot^2(x)-\frac{1}{3}\cos(x)\cot^2(x)\]\[=\frac{1}{6}\cos(x)\cot^2(x).\]
If it is not right, then I may have made a sign error, but you get the idea.
but the answer is wrong :(
you said "y=C1cosx+C2sinx+1/6cotxcscx that's my answer but i dont know where does 1/6cotxcscx came from. :)" but we are getting y=C1cosx+C2sinx+1/6cot^2x*cscx (across and I get the exact same result) so I am betting on a typo somewhere perhaps
oh maybe they just simplfy it more? but our answer at first is correct? right?
\[\frac16\cot^2x\csc x\neq\frac16\cot x\csc x\]so it's not a simplification issue
Since across and I get the exact same result I am going to say that we have provided the correct answer to the posted problem. Be sure the problem as you posted it is free of typos
hmmm.. what do u mean typos? :)
are you \(sure\) the original problem is\[y''+y=\csc^3x\cot x\]? If you wrote it wrong, that would be a "typo" I think the problem is written correct though, because this makes the integration easy, which means there is likely a typo in the solution in the back of your book; i.e. it should be cot^2x not cotx (an easy typo for a book to make)
ahh... yep that's the problem. hmm. so which answer should i follow? YOURS? or THE BOOK? HAHA
Just talk to your professor if you seek further clarification.
hmmm.. but this is a bring home exam. i hope he will tell me that the book is wrong HAHA.. i think i will follow your process. and i will consult also my prof tom. thank you for solving it. i hope i can count on you next time :)
Across and I came to the same answer individually. I'm self-taught in DE's and make mistakes at times, but across goes to MIT, so I'm pretty sure the book is wrong
and it looks like wolfram agrees with across and I, so I'm going to say book is wrong: final answer http://www.wolframalpha.com/input/?i=y%22%2By%3Dcsc%5E3xcotx
HAHAHA. great! THANKS to you two :)
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