Question

(e^(-x^2))dy/x + dx/cos^2y=0

Answer 1

$\frac{ {e}^{ - {x}^{2} } dy}{x} + \frac{dx}{ \cos {}^{2} (y) } = 0 \\ \frac{ {e}^{ - {x}^{2}}dy }{x} = - \frac{dx}{ \cos {}^{2} (y) } \\ \int\limits { \cos }^{2} y dy= - \int\limits \frac{xdx}{e {}^{ - {x}^{2} } } \\ \int\limits \frac{1 + \cos(2y) }{2} dy = - \int\limits {e}^{ {x}^{2} }x dx \\ \frac{1}{2} \int\limits(1 + \cos(2y)) dx = - \frac{1}{2} \int\limits {e}^{ {x}^{2} } 2xdx \\ \frac{1}{2} (\int\limits \: dx + \frac{1}{2}\int\limits \cos(2y)d(2y)) = - \frac{1}{2} \int\limits {e}^{ {x}^{2} } d( {x}^{2} ) \\ \frac{1}{2} (y + \frac{1}{2} \sin(2y)) = - \frac{1}{2} {e}^{ {x}^{2} } + c \\ y + \frac{1}{2} \sin(2y) = - {e}^{ {x}^{2} } + c$

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