trigonometric equation

**transgalactic** · November 11th 2010, 01:09 PM

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

**pickslides** · November 11th 2010, 01:20 PM

Originally Posted by transgalactic

B)
$x^{2}y'-1=cos2y$

Does this separate?

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

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

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

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

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

It does! what next?

**transgalactic** · November 13th 2010, 05:28 AM

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

regarding A)
i put v=cosy

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

**transgalactic** · November 17th 2010, 01:01 AM

how to solve A

**Ackbeet** · November 17th 2010, 02:43 AM

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

**transgalactic** · November 17th 2010, 03:02 AM

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

what to do now?

**Ackbeet** · November 17th 2010, 04:48 AM

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

**transgalactic** · November 17th 2010, 06:44 AM

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

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

i dont know how to solve the right side integral

**Ackbeet** · November 17th 2010, 07:21 AM

Original DE:

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

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

Plugging into the DE yields

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

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

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

**transgalactic** · November 18th 2010, 10:57 AM

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

i dont know how to solve this integral

**Ackbeet** · November 18th 2010, 11:01 AM

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

**TheCoffeeMachine** · November 18th 2010, 11:41 AM

Originally Posted by transgalactic

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

$\frac{1}{\cos{2y}+1} = \frac{1}{2}\sec^2{y}$ which is the derivative of $\frac{1}{2}\tan{y}$ .

**transgalactic** · November 18th 2010, 11:42 AM

thanks

**Ackbeet** · November 18th 2010, 12:18 PM

So, did you finish?

**transgalactic** · November 18th 2010, 12:25 PM

yes

i solved just like you said it went perfect

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