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I was helping somebody on some math problems when a wild question appears. It goes like this:

Get the solution to the differential equation $$\cos y \sin2x dx +(\cos^2y - \cos^2x)dy = 0$$

My work

I was able to search the internet on how to get the solution to the differential equation shown above but I do not understand the way.

We could perhaps rearrange the above differential equation into the form

$$M(x,y)dx + N(x,y)dy = 0$$

With the form above, we could see if the given differential equation is variable-separable, homogenous or an exact-type.

Testing the given differential equation as variable-separable:

No. We can't. The differential equation above isn't variable-separable.

Testing the given differential equation as homogenous:

$F(x,y)$ is a homogenous function of degree $n$ in $x$ and $y$ provided $F(kx,ky)$ = $k^nF(x,y)$. For example, $x^2 + y^2$ and $x^2 \sin \left(\frac{y}{x}\right)$ are homogenous functions of degree $2$ in $x$ and $y$. With that in mind...

Looking at the first term $\cos y \sin2x$: $$F(x,y) = \cos y \sin2x$$ $$F(kx,ky) = \cos (ky) \sin(2kx)$$ Getting deeper, we conclude that $$F(kx,ky) \neq k^nF(x,y)$$

Looking at the second term $\cos^2y - \cos^2x$: $$F(x,y) = \cos^2y - \cos^2x$$ $$F(kx,ky) = \cos^2(ky) - \cos^2(kx)$$ Getting deeper, we conclude that $$F(kx,ky) \neq k^nF(x,y)$$

The terms themselves must be homogenous if the differential equations are homogenous.

No. We can't. The differential equation above isn't homogenous.

Testing the given differential equation if it is an exact-type:

A necessary condition for a differential equation to be exact is

$$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$

So getting the expression for $\frac{\partial M}{\partial y}$:

$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(\cos y \sin2x) $$ $$\frac{\partial M}{\partial y} = \sin 2x\frac{\partial}{\partial y}(\cos y) $$ $$\frac{\partial M}{\partial y} = \sin 2x(-\sin y)$$ $$\frac{\partial M}{\partial y} = -\sin y \sin 2x$$

So getting the expression for $\frac{\partial N}{\partial x}$:

$$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(\cos^2y - \cos^2x) $$ $$\frac{\partial N}{\partial x} = 0 - \frac{\partial}{\partial x}(\cos^2x) $$ $$\frac{\partial N}{\partial x} = 0 - \frac{\partial}{\partial x}\left(\frac{1}{2} + \frac{1}{2}\cos 2x\right) $$ $$\frac{\partial N}{\partial x} = \sin 2x $$

We see that $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, so the given differential equation isn't exact either.

Let's try converting it into linear differential equation.

Let's try rearranging the given differential equation above into a Bernoulli equation, having the form

$$\frac{dy}{dx} + P(x)y = Q(x)$$ Then...

$$\cos y \sin2x dx +(\cos^2y - \cos^2x)dy = 0$$ $$\frac{\cos y \sin2x dx}{dx} +\frac{(\cos^2y - \cos^2x)dy}{dx} = \frac{0}{dx}$$ $$\cos y \sin2x + (\cos^2y - \cos^2x)\frac{dy}{dx} = 0$$ $$\frac{\cos y \sin2x}{\cos^2y - \cos^2x} + \frac{\cos^2y - \cos^2x}{\cos^2y - \cos^2x}\frac{dy}{dx} = \frac{0}{\cos^2y - \cos^2x}$$ $$\frac{\cos y \sin2x}{\cos^2y - \cos^2x} + \frac{dy}{dx} = 0$$ $$\frac{dy}{dx} + \frac{\cos y \sin2x}{\cos^2y - \cos^2x} = 0$$

Screw that. We couldn't even discern what is the expression of $P(x)y$ because the expression $\frac{\cos y \sin2x}{\cos^2y - \cos^2x}$ is irreducible. We can't separate $x$'s from $y$'s.

How do we get the solution of the differential equation above?

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    $\begingroup$ I may suggest using substitutions: $X=sin(x)$ and $Y=cos(y)$. However, after some work, I don’t think that helps much. In general, I think there is no closed form solution for this equation. This can be solved by some algorithms like Picard’s method. $\endgroup$ – Szeto Feb 14 '18 at 15:43
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Just a little hint

This is not a full solution ...

$$\cos(y)\sin(2x)x'=-\cos^2(y)+ {\cos^2(x)}$$ $$\sin(2x)x'=-\cos(y)+\frac {\cos^2(x)}{\cos(y)}$$ $$2\sin(x)\cos(x)x'=-\cos(y)+\frac {1-\sin^2(x)}{\cos(y)}$$ $$2\sin(x)\cos(x)x'+\frac {\sin^2(x)}{\cos(y)}=-\cos(y)+\frac {1}{\cos(y)}$$ $$2\sin(x)\cos(x)x'+\frac {\sin^2(x)}{\cos(y)}=f(y)$$ $$2\sin(x)(\sin(x))'+\frac {\sin^2(x)}{\cos(y)}=f(y)$$ Sustitute $h=\sin(x)$ for a better expression $$2hh'+\frac {h^2}{\cos(y)}=f(y)$$ $$(h^2)'+\frac {h^2}{\cos(y)}=f(y)$$ Substitute $g(y)=h^2(y)-1$ $$\boxed{g'(y)+\frac {g(y)}{\cos(y)}=-\cos(y)}$$

$$........$$ We have reduced the first equation to the canonical form of a first order linear equation. But it does not mean that it's easy to integrate such ode. Especially if it involves Gaussian like integrals...

Edit Maybe the ODE could be solved with $tan(x/2)$ susbtitution's formula... when I write $h=\sin(x)$ it means $h(y)=\sin(x(y))$ both x, and h are functions of the variable y.And so is g ( g(y)) too.

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    $\begingroup$ $h = \sin(x)$ and $g = h^2 = \sin^2(x)$, then what is $g(y)$? $\endgroup$ – AlexanderJ93 Feb 14 '18 at 16:55
  • $\begingroup$ g is a function of y ......because I considered x as the function and y as the variable @AlexanderJ93 and $\sin(x)$ is a fucntion of the variable y Note in the first lines I considered x' and not y' $\endgroup$ – Isham Feb 14 '18 at 16:58
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    $\begingroup$ I see. Thanks for clarifying. $\endgroup$ – AlexanderJ93 Feb 14 '18 at 16:59
  • $\begingroup$ I don't understand....Why from this expression $(h^2)'+\frac {h^2}{\cos(y)}=f(y)$, when we let $g(y)=h^2(y)-1$, it instantly becomes $\boxed{g'(y)+\frac {g(y)}{\cos(y)}=-\cos(y)}$? $\endgroup$ – Palautot Ka Feb 18 '18 at 2:53
  • $\begingroup$ @PalautotKa Look at the value of f(y) .$$f(y)=-\cos(y)+\frac {1}{\cos(y)}$$ I just use that for simplicity ..thats all $$(h^2)'+\frac {h^2}{\cos(y)}=-\cos(y)+\frac {1}{\cos(y)}$$ $$(h^2)'+\frac {h^2-1}{\cos(y)}=-\cos(y)$$$$g'+\frac {g}{\cos(y)}=-\cos(y)$$ $\endgroup$ – Isham Feb 18 '18 at 3:06
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Sorry, just now I do not have the time to type the answer in latex. I'll do it tomorrow if it appears useful.

This is only a quick overview, without care about the signs of the square roots. Hoping it will be an useful hint.

enter image description here

The solution is on the form $x(y)$. Probably, there is no closed form for the inverse function $y(x)$.

Note :

$(1+\sin(y))\cos^2(x)=c\:\cos(y)-cos^2(y)-\cos(y)\sin^{-1}(\cos(y))$

$\frac{(1+\sin(y))\cos^2(x)}{\cos(y)}=c-\cos(y)-\sin^{-1}(\cos(y))=C-\cos(y)+y$

$$y-\cos(y)-\frac{(1+\sin(y))\cos^2(x)}{\cos(y)}=-C$$ This is consistent with the expected solution (doesn't matter the sign of $C$).

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  • $\begingroup$ I admired your effort.....I can wait for your answer. Just take it easy. No rush.....:-) $\endgroup$ – Palautot Ka Feb 15 '18 at 2:27
  • $\begingroup$ The solution of the differential equation I've seen on Internet is $y−\cos (y)−\cos ^2(x)\frac{1+sin(y)}{cos(y)}=C$. They got the solution because they made the given differential equation, which is an inexact d.e, into an exact d.e. I don't know if the solution is right. Maybe we can work it out..... $\endgroup$ – Palautot Ka Feb 15 '18 at 2:33
  • $\begingroup$ The solutions are both correct, it's just that @JJacquelin integrated $$\int\frac{-dt}{\sqrt{1-t^2}}=c-\sin^{-1}t$$ rather than $$\int\frac{-dt}{\sqrt{1-t^2}}=\cos^{-1}t-C$$ $\endgroup$ – user5713492 Feb 15 '18 at 4:48
  • $\begingroup$ @Palautot Ka: The above solution is consistent with the solution you found on internet (See the added note at end of my answer). So, no need to waste time to re-edit my first answer. $\endgroup$ – JJacquelin Feb 17 '18 at 10:58

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