Question

dy/dx=(1+2x^2)/(x cos y)

Answer 1

What's the question?

Answer 2

Find the general solution to the following differential equation :

Answer 3

http://www.wolframalpha.com/input/?i=dy%2Fdx%3D%281%2B2x%5E2%29%2F%28x+cos+y%29

Answer 4

\[\cos(y) dy=\frac{1+2x^2}{x} dx\]
integrate both sides

Answer 5

\[\int\limits_{}^{}\cos(y) dy=\int\limits_{}^{}(\frac{1}{x}+2x) dx\]

Answer 6

\[\sin(y)=\ln|x|+x^2+C\]

Answer 7

∫ (1+2x^2)/(x cos y) dx=( X^2+log(x))sec(y)+c
Is this correct myininaya ?

Answer 8

we have to do separation of variables

Answer 9

we had
\[\frac{dy}{dx}=\frac{1+2x^2}{x \cos(y)}\]
I want to get my y's together and my x's together
so first thing i did was multiply cos(y) on both sides
\[\cos(y) \frac{dy}{dx}=\frac{1+2x^2}{x}\]
Then I multiplied dx on both sides
\[\cos(y) dy=\frac{1+2x^2}{x} dx\]

Answer 10

Now we can integrate both sides

Answer 11

i have to say thank you as well.

Answer 12

Aww... you're welcome :)