MaxWong

Oh Thanks.

Also:

Real number: \Bbb{R}  \(\Bbb{R}\)

Rational number: \Bbb{Q}   \(\Bbb{Q}\)

Integers: \Bbb{Z}   \(\Bbb{Z}\)

Natural number: \Bbb{N}   \(\Bbb{N}\)

Complex number: \Bbb{C} \(\Bbb{C}\)

Oh thank you.

I am not very careful

OK. I thought it was correct...... Let me do it myself......

\(2xy\text{ dx} + (1+x^2) \text{ dy} = 0\\ 2xy\text{ dx} = -(1+x^2) \text{ dy}\\ -\dfrac{2x}{1+x^2} \text{ dx} = \dfrac{\text{ dy}}{y}\\ \displaystyle\int -\dfrac{2x}{1+x^2} \text{ dx} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ \text{ Let }u = 1 + x^2,\text{ du} = 2x \text{ dx}\\ -\displaystyle\int \dfrac{\text{ du}}{u} = \displaystyle\int\dfrac{\text{ dy}}{y}\\ -\ln |u| = \ln |y| + C\\ \ln |\dfrac{1}{u}| = \ln |y| + C\\ \ln |\dfrac{1}{1+x^2}|= \ln |y| + C\\ \text{ Set }C = \ln K\\ \ln (\dfrac{1}{1+x^2})=\ln |y + K|\\ |y + K| = \dfrac{1}{1+x^2}\\ y + K = \pm \dfrac{1}{1+x^2}\\ y = K\pm\dfrac{1}{1+x^2}\)

I looked up the Internet and this may be useful.

Happy New Year!!

Didn't even notice it is already 2017 until I see this post :P

Too focused on Maths recently :D

It is Okay you don't know how to use LaTeX, you still have 2 ways to give me the answer......

1) Do the LaTeX with assistance of my LaTeX tutorial posts in the 'Sticky Topics'

2) Write it on a paper, Take a photo, import in computer, and use the Image function of the forum to import the photo into the post.

The answer is correct.

Btw Happy New Year :)

Also this awesome derivative calculator here!!

I did not pay for the W|A so I do not have access.

LOLOLOLOL :P