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Question Number 149757 by iloveisrael last updated on 07/Aug/21

Commented by amin96 last updated on 07/Aug/21
$x(dy/dx)+2y=((sinx)/x) (dy/dx)=((sin x−2xy)/x^2 ) (sin x−2xy)dx+(−x^2 )dy=0 (∂M/∂y)=−2x (∂N/∂x)=−2x M_y =N_x ∫Mdx=∫Ndy ⇒ { ((∫Mdx=−cos x−x^2 y+C(y))),((∫Ndy=−x^2 y+C(x))) :} { ((C(y)=0)),((C(x)=−cos x )) :} C=−cos x−x^2 y y=((c+cos x)/(−x^2 ))$
$x \frac{d y}{d x} + 2 y = \frac{s i n x}{x} \frac{d y}{d x} = \frac{\sin x - 2 x y}{x^{2}}$ $(\sin x - 2 x y) d x + (- x^{2}) d y = 0$ $\frac{\partial M}{\partial y} = - 2 x \frac{\partial N}{\partial x} = - 2 x M_{y} = N_{x}$ $\int M d x = \int N d y \Rightarrow {\begin{cases} \int M d x = - \cos x - x^{2} y + C (y) \\ \int N d y = - x^{2} y + C (x) \end{cases}$ ${\begin{cases} C (y) = 0 \\ C (x) = - \cos x \end{cases} C = - \cos x - x^{2} y$ $y = \frac{c + \cos x}{- x^{2}}$
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