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anonymous 4 years ago dy/dx=(1+2x^2)/(x cos y)

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1. anonymous

What's the question?

2. anonymous

Find the general solution to the following differential equation :

3. anonymous
4. myininaya

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

5. myininaya

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

6. myininaya

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

7. anonymous

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

8. myininaya

we have to do separation of variables

9. myininaya

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 $\cos(y) dy=\frac{1+2x^2}{x} dx$

10. myininaya

Now we can integrate both sides

11. anonymous

i have to say thank you as well.

12. myininaya

Aww... you're welcome :)

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