A)

$\displaystyle y'siny=cosx(2cosy-sin^{2}x)$

B)

$\displaystyle x^{2}y'-1=cos2y $

how to solve these

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- Nov 11th 2010, 01:09 PMtransgalactictrigonometric equation
A)

$\displaystyle y'siny=cosx(2cosy-sin^{2}x)$

B)

$\displaystyle x^{2}y'-1=cos2y $

how to solve these - Nov 11th 2010, 01:20 PMpickslides
- Nov 13th 2010, 05:28 AMtransgalactic
what is the integral of 1/(cos2y+1)

?

regarding A)

i put v=cosy

$\displaystyle -v'=cosx(2v-sin^2x)$ - Nov 17th 2010, 01:01 AMtransgalactic
how to solve A

- Nov 17th 2010, 02:43 AMAckbeet
For (A), the substitution $\displaystyle u=\cos(y)$ renders the equation linear.

- Nov 17th 2010, 03:02 AMtransgalactic
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? - Nov 17th 2010, 04:48 AMAckbeet
You did the substitution incorrectly. u = sin(y). sin(x) remains as is.

- Nov 17th 2010, 06:44 AMtransgalactic
i got

u'+2ucosx=-(sin x)^2

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

i dont know how to solve the right side integral - Nov 17th 2010, 07:21 AMAckbeet
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? - Nov 18th 2010, 10:57 AMtransgalactic
on the left side i get cos x (sin x)^2 e^{2sinx}

i dont know how to solve this integral - Nov 18th 2010, 11:01 AMAckbeet
Substitute v = sin(x). Then you'll need by parts twice.

- Nov 18th 2010, 11:41 AMTheCoffeeMachine
- Nov 18th 2010, 11:42 AMtransgalactic
thanks :)

- Nov 18th 2010, 12:18 PMAckbeet
So, did you finish?

- Nov 18th 2010, 12:25 PMtransgalactic
yes

:)

i solved just like you said it went perfect