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genius12 Group TitleBest ResponseYou've already chosen the best response.3
are you supposed to find dy/dx?
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
im not confident of the answer in my book
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
Just rearrange. Take the 2dy to the other side, divide both sides by dx, and then divide both sides by 2.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
can you show solution??
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
First add 2dy to both sides:\[\bf xy^3\ln(x)dx2dy=0 \implies xy^3\ln(x)dx=2dy\]Now divide both sides by dx:\[\bf xy^3\ln(x)=2\frac{dy}{dx}\]Now divide both sides by 2:\[\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\]
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
variable seperable method?
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
@niah2411 Do you see what I did?
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
maybe my book is wrong
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
add 2dy to both sides then divide both sides by 2dx.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
our methods are the same and also the answers, the book answer is maybe wrong
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
May be. It often happens that the publishers make a mistake in the solutions.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
the book answer is x^2(lnx^21)+4y^2=c
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
That looks like you're supposed to find the solution to the equation, then, not just find dy/dx
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
Can you stop tagging everyone. Please don't pay attention guys.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
@agent0smith can you give me a solution if you have one?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
Sorry, don't really remember much of this part of calculus :(
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
@genius12 can you give me a solution using variable separable method?
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
oh..i see i understand that
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
Oh, yes you can use separation of variables: \[\Large \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}\] multiply both sides by dx, divide both sides by y^3\[\Large \int\limits \frac{ x\ln(x) }{ 2 } dx= \int\limits \frac{dy}{y^3}\]Now you can integrate.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
ohh can you give me your full solution so i can understand it really..pls?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
Make it a little simpler to integrate:\[\Large \frac{ 1 }{ 2 } \int\limits\limits x \ln(x) dx= \int\limits\limits y^{3} dy\] On the left it looks like you'll need to use integration by parts. Sorry, I'm too tired to go through it all now, someone else can take over.
 one year ago

niah2411 Group TitleBest ResponseYou've already chosen the best response.0
oh im sorry @agent0smith for the inconvenience
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
@niah2411 I'll take over from where @agent0smith left off. To continue, we will use integration by parts for the "xln(x)" integral: \[\bf \frac{ 1 }{ 2 }\int\limits_{}^{}xln(x) \ dx \implies \frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}\int\limits_{}^{}\frac{ x }{ 2} \ dx \right)=\frac{ 1 }{ 2 } \left( \frac{x^2\ln(x)}{2}\frac{ x^2 }{ 4} \right)\]\[\bf =\frac{ 2x^2\ln(x)x^2 }{ 8 }+C\]Now the rightside would just be:\[\bf \int\limits_{}^{}y^{3} \ dy=\frac{y^{2}}{2}\]Now we equate the results: \[\bf \frac{1}{2y^2}=\frac{ 2x^2\ln(x)x^2 }{ 8 }+C\]
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
Can you solve for 'y' now?
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
@agent0smith You still here? Help me awaken @niah2411 ? lol
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
Hehe yeah I'm still here. After saying i was too tired, i did the int by parts on paper (i thought it would take longer than it did, or i would've just typed it up). Looks like he didn't need to find y anyway, this was the answer: x^2(lnx^21)+4y^2=c which you can get by moving terms around and multiplying by 8.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
@agent0smith I don't know what you're saying?
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
This was the answer that he gave, from the book: x^2(lnx^21)+4y^2=c Which you can get from your result.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.3
According to my result, which is correct, the answer should be:\[\bf x^2(2\ln(x)1)+4y^{2}=C\]So yes, the book answer is correct and I believe he misinterpreted the meaning of "solution" or whatever. They were just looking for separation of variables and then simply bring over the x's and y's' on one side and equating it to the constant. @agent0smith
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.3
Yep, that's what i worked out. The book/yours is the same.
 one year ago
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