## niah2411 one year ago differential equation of xy^3lnxdx-2dy=0

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1. genius12

are you supposed to find dy/dx?

2. niah2411

yup..

3. niah2411

im not confident of the answer in my book

4. genius12

Just re-arrange. Take the -2dy to the other side, divide both sides by dx, and then divide both sides by 2.

5. niah2411

can you show solution??

6. genius12

First add 2dy to both sides:$\bf xy^3\ln(x)dx-2dy=0 \implies xy^3\ln(x)dx=2dy$Now divide both sides by dx:$\bf xy^3\ln(x)=2\frac{dy}{dx}$Now divide both sides by 2:$\bf \therefore \ \frac{ xy^3\ln(x) }{ 2 }=\frac{dy}{dx}$

7. niah2411

variable seperable method?

8. genius12

@niah2411 Do you see what I did?

9. niah2411

yes

10. niah2411

maybe my book is wrong

11. genius12

add 2dy to both sides then divide both sides by 2dx.

12. niah2411

our methods are the same and also the answers, the book answer is maybe wrong

13. genius12

May be. It often happens that the publishers make a mistake in the solutions.

14. niah2411

15. agent0smith

That looks like you're supposed to find the solution to the equation, then, not just find dy/dx

16. genius12

Can you stop tagging everyone. Please don't pay attention guys.

17. niah2411

@agent0smith can you give me a solution if you have one?

18. agent0smith

Sorry, don't really remember much of this part of calculus :(

19. niah2411

@genius12 can you give me a solution using variable separable method?

20. niah2411

oh..i see i understand that

21. agent0smith

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.

22. niah2411

ohh can you give me your full solution so i can understand it really..pls?

23. agent0smith

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.

24. niah2411

oh im sorry @agent0smith for the inconvenience

25. genius12

@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$Now the right-side would just be:$\bf \int\limits_{}^{}y^{-3} \ dy=-\frac{y^{-2}}{2}$Now we equate the results: $\bf -\frac{1}{2y^2}=\frac{ 2x^2\ln(x)-x^2 }{ 8 }+C$

26. genius12

Can you solve for 'y' now?

27. genius12

@agent0smith You still here? Help me awaken @niah2411 ? lol

28. agent0smith

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^2-1)+4y^-2=c which you can get by moving terms around and multiplying by 8.

29. genius12

@agent0smith I don't know what you're saying?

30. agent0smith

This was the answer that he gave, from the book: x^2(lnx^2-1)+4y^-2=c Which you can get from your result.

31. genius12

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

32. agent0smith

Yep, that's what i worked out. The book/yours is the same.