# Thread: trigonometric equation

1. ## trigonometric equation

A)
$\displaystyle y'siny=cosx(2cosy-sin^{2}x)$
B)
$\displaystyle x^{2}y'-1=cos2y$
how to solve these

2. Originally Posted by transgalactic
B)
$\displaystyle x^{2}y'-1=cos2y$
Does this separate?

$\displaystyle x^{2}y'-1=\cos 2y$

$\displaystyle x^{2}y'=\cos 2y+1$

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

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

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

It does! what next?

3. what is the integral of 1/(cos2y+1)
?

regarding A)
i put v=cosy

$\displaystyle -v'=cosx(2v-sin^2x)$

4. how to solve A

5. For (A), the substitution $\displaystyle u=\cos(y)$ renders the equation linear.

6. from this substitution i got
$\displaystyle -u'=2u^2 -u(sin x)^2$
by rearanging i got
u'/u +2u=(sin x)^2

what to do now?

7. You did the substitution incorrectly. u = sin(y). sin(x) remains as is.

8. i got
u'+2ucosx=-(sin x)^2

$\displaystyle (ue^{2sinx})'=(sin x)^2e^{2sinx}$

i dont know how to solve the right side integral

9. Original DE:

$\displaystyle y'\sin(y)=\cos(x)(2\cos(y)-\sin^{2}(x)).$

Substitution: $\displaystyle u=\cos(y).$ Then $\displaystyle -u'=\sin(y)y'.$

Plugging into the DE yields

$\displaystyle -u'=\cos(x)(2u-\sin^{2}(x))\quad\to\quad -u'-2u\cos(x)=-\cos(x)\sin^{2}(x)$

$\displaystyle \to\quad u'+2u\cos(x)=\cos(x)\sin^{2}(x).$

I think you forgot to distribute the cosine everywhere. So, propagate these changes: what do you get?

10. on the left side i get cos x (sin x)^2 e^{2sinx}

i dont know how to solve this integral

11. Substitute v = sin(x). Then you'll need by parts twice.

12. Originally Posted by transgalactic
what is the integral of 1/(cos2y+1)?
$\displaystyle \frac{1}{\cos{2y}+1} = \frac{1}{2}\sec^2{y}$ which is the derivative of $\displaystyle \frac{1}{2}\tan{y}$.

13. thanks

14. So, did you finish?

15. yes

i solved just like you said it went perfect

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