$\left(x^2 - \sin ^2\left(y\right)\right)dx + x\sin \left(2y\right)dy = 0$, $y\left(1\right) = \frac{π}{3}$. $x = ?$

As my final answer, I get: $x=\pm\sqrt{\frac{7}{4}\frac{1}{\sin^2\left(y\right)}}$, but it tells me that it is wrong. I've tried it multiple times and get the same thing. Anyone help?

$M_y = -2\sin \left(y\right)\cos \left(y\right)$ and $N_x = 2\sin \left(y\right)\cos \left(y\right)$. $M_y ≠ N_x$.

$\frac{M_y - N_x}{N} = \frac{-2\sin \left(y\right)\cos \left(y\right) - 2\sin \left(y\right)\cos \left(y\right)}{x2\sin \left(y\right)\cos \left(y\right)} = \frac{-4\sin \left(y\right)\cos \left(y\right)}{x2\sin \left(y\right)\cos \left(y\right)} = \frac{-2}{x}$

Then, $u(x) = e^{\int \frac{M_y - N_x}{N}} = e^{\int \frac{-2}{x}} = e^{-2 \ln(x)} = x^{-2}$ (we can leave off the constant for now)

Then, multiplying through by $u\left(x\right)$, we have: $x^{-2}\left(x^2-\sin ^2\left(y\right)\right)dx + x^{-1}\sin \left(2y\right)dy = 0$. $M_y = -x^{-2}2\sin \left(y\right)\cos \left(y\right)$, $N_x = -x^{-2}2\sin \left(y\right)\cos \left(y\right)$.

$M_y = N_x$

I ended up getting $x = \pm \sqrt{\frac{7}{4}\frac{1}{\sin ^2\left(y\right)} }$, but it tells me i am wrong.

Resolved: I figured it out*


You've miscalculated. $$\frac{-4\sin y\cos y}{x\cdot2\sin y\cos y}=\frac{-2\cdot2\sin y\cos y}{x\cdot 2\sin y\cos y}=\frac{-2}{x},$$ so long as $\sin y\cos y\ne 0$ and $x\ne 0.$


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