Question

how to find the general solution of this:
cos y sin 2x dx + ( (cos y)^2 -(cos x )^2)dy = 0

I am trying substitution but I can't find anything that will lead me to a solution.

Answer 1

anyone?

Answer 2

yes

Answer 3

the answer is:
[cos x(1 + siny)]^2 = cos y(y+c-cos y)
I don't know how to get there

Answer 4

sorry but I can't understand your solution
could you explain it further?
thanks :)

Answer 5

isn't it that you can't integrate it immediately ?

Answer 6

\[\cos y\sin2x~dx+\left(\cos^2y-\cos^2x\right)~dy=0\]
Check to see if it's exact:
\[\begin{cases}M(x,y)=\cos y\sin2x\\N(x,y)=\cos^2y-\cos^2x\end{cases}~~\Rightarrow~~\begin{cases}\dfrac{\partial M}{\partial y}=-\sin y\sin 2x\\\\
\dfrac{\partial N}{\partial x}=2\cos x\sin x=\sin2x
\end{cases}\]
the equation isn't exact, but we can find an integrating factor in terms of \(y\):
\[\begin{align*}\ln\mu(y)&=\int\frac{\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}}{M}~dy\\\\
&=\int\frac{\sin2x+\sin y\sin2x}{\cos y\sin2x}~dy\\\\
&=\int\frac{1+\sin y}{\cos y}~dy\\\\
&=\int(\sec y+\tan y)~dy\\\\
&=\ln|\sec y+\tan y|-\ln|\cos y|\\\\
&=\ln\left|\frac{\sec y+\tan y}{\cos y}\right|\\\\
\mu(y)&=\frac{\sec y+\tan y}{\cos y}\\\\
&=\frac{1+\sin y}{\cos^2y}
\end{align*}\]Now you should have an exact equation:
\[\begin{align*}\frac{1+\sin y}{\cos^2y}\cos y\sin2x~dx+\frac{1+\sin y}{\cos^2y}\left(\cos^2y-\cos^2x\right)~dy&=0\\\\
\frac{\sin2x+\sin2x\sin y}{\cos y}~dx+\left(1+\sin y-\frac{1+\sin y}{\cos^2y}\cos^2x\right)~dy&=0
\end{align*}\]
So now,
\[\begin{cases}M=\dfrac{\sin2x+\sin2x\sin y}{\cos y}\\\\N=1+\sin y-\dfrac{1+\sin y}{\cos^2y}\cos^2x\end{cases}~~\Rightarrow~~\begin{cases}\dfrac{\partial M}{\partial y}=\dfrac{\sin2x(1+\sin y)}{\cos^2y}\\\\
\dfrac{\partial N}{\partial x}=\dfrac{\sin2x(1+\sin y)}{\cos^2y}
\end{cases}\]
which means the equation is exact.

Answer 7

We're looking for a solution of the form \(\Psi(x,y)=C\), where \(\dfrac{\partial \Psi}{\partial x}=M(x,y)\) and \(\dfrac{\partial \Psi}{\partial y}=N(x,y)\).
\[\begin{align*}
\dfrac{\partial\Psi}{\partial x}&=M\\\\
\partial\Psi&=\frac{(1+\sin y)\sin2x}{\cos y}~\partial x\\\\
\int\partial\Psi&=\frac{1+\sin y}{\cos y}\int\sin2x~\partial x\\\\
\color{red}\Psi&\color{red}{=-\frac{(1+\sin y)\cos2x}{2\cos y}+f(y)}\\\\
\frac{\partial\Psi}{\partial y}&=-\frac{1}{2}\frac{\partial}{\partial y}\left[\frac{(1+\sin y)\cos2x}{\cos y}\right]+f'(y)\\\\
N(x,y)&=-\frac{\cos2x(1+\sin y)}{2\cos^2y}+f'(y)\\\\
1+\sin y-\dfrac{1+\sin y}{\cos^2y}\cos^2x&=-\frac{\cos2x(1+\sin y)}{2\cos^2y}+f'(y)\\\\
f'(y)&=(1+\sin y)\left(1-\frac{\cos^2x}{\cos^2y}+\frac{\cos2x}{2\cos^2y}\right)\\\\
f'(y)&=(1+\sin y)\left(1-\frac{1+\cos2x}{2\cos^2y}+\frac{\cos2x}{2\cos^2y}\right)\\\\
f'(y)&=(1+\sin y)\left(1-\frac{1}{2\cos^2y}\right)\\\\
f'(y)&=1+\sin y-\frac{1}{2}\sec^2y-\frac{1}{2}\frac{\sin y}{\cos^2y}\\\\
f(y)&=\int\left(1+\sin y-\frac{1}{2}\sec^2y-\frac{1}{2}\sec y\tan y\right)~dy\\\\
\color{blue}{f(y)}&\color{blue}{=y-\cos y-\frac{1}{2}\tan y-\frac{1}{2}\sec y+C}
\end{align*}\]
So the final solution should be
\[\Psi(x,y)=-\frac{(1+\sin y)\cos2x}{2\cos y}+y-\cos y-\frac{1}{2}\tan y-\frac{1}{2}\sec y+C=0\]
or simply
\[-\frac{(1+\sin y)\cos2x}{2\cos y}+y-\cos y-\frac{1}{2}\tan y-\frac{1}{2}\sec y=C\]