| PRZYKŁAD 1 |
2xtgy + (x2−2siny)y′ = 0
$$2x\text{tg}y + \left( x^{2} - 2\sin y \right)\frac{\text{dy}}{\text{dx}} = 0\ \ / \bullet dx$$
2xtgydx+(x2−2siny)dy = 0
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P(x, y)=2xtgy , Q(x, y) = x2 − 2siny
$$\frac{\mathbf{\partial P}}{\mathbf{\partial y}} = \left( 2x\text{tg}y \right)^{'} = \mathbf{2}\mathbf{x \bullet}\frac{\mathbf{1}}{\mathbf{\cos}^{\mathbf{2}}\mathbf{y}}\text{\ \ \ \ \ \ \ \ \ \ \ \ }\frac{\mathbf{\partial Q}}{\mathbf{\partial x}} = \left( x^{2} - 2\sin y \right)^{'} = \mathbf{2}\mathbf{x}\ $$
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$$\frac{\mathbf{\partial P}}{\mathbf{\partial y}}\mathbf{\neq}\frac{\mathbf{\partial Q}}{\mathbf{\partial x}}$$
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$\mathbf{u}\left( \mathbf{x} \right) \Longrightarrow \frac{1}{x^{2} - 2\sin y}\left( \frac{2x}{\cos^{2}y} - 2x \right)$
$$\mathbf{u}\left( \mathbf{y} \right) \Longrightarrow \frac{1}{2x\text{tgy}}\left( 2x - \frac{2x}{\cos^{2}y} \right) = \frac{2x}{2xtgy} - \frac{2x}{2x\cos^{2}y\text{tg}y} =$$
$$= \frac{\cos^{2}y - 1}{\text{tg}y\cos^{2}y} = \frac{- \sin^{2}y}{\frac{\sin y}{\cos y} \bullet \cos^{2}y} = \frac{- \sin^{2}y}{\sin y\cos y} = - \frac{\sin y}{\cos y} = \mathbf{-}\mathbf{\text{tg}}\mathbf{y}$$
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u(y) = e−∫tgy = eln|cosy| = cosy
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cosy2xtgydx+cosy(x2−2siny)dy = 0
2xsinydx + (x2cosy−sin2y)dy = 0
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P(x, y) = 2xsiny Q(x, y) = x2cosy − sin2y
$$\frac{\mathbf{\partial P}}{\mathbf{\partial y}} = \left( 2x\sin y \right)^{'} = \mathbf{2}\mathbf{x}\cos\mathbf{y}\mathbf{\ }\text{\ \ \ \ }\frac{\mathbf{\partial Q}}{\mathbf{\partial x}} = \left( x^{2}\cos y - \sin{2y} \right)^{'} = \mathbf{2}\mathbf{x}\cos\mathbf{y}$$
$$\frac{\mathbf{\partial P}}{\mathbf{\partial y}}\mathbf{=}\frac{\mathbf{\partial Q}}{\mathbf{\partial x}}$$
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S(0, 0)
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F(x, y) = ∫0x2tsinydt + ∫0y(−sin2t)dt=
$$= 2\sin y \bullet \left. \ \frac{1}{2}t^{2} \right|\begin{matrix}
x \\
0 \\
\end{matrix} + \left. \ \frac{1}{2}\cos{2t} \right|\begin{matrix}
y \\
0 \\
\end{matrix} = x^{2}\sin y + \frac{1}{2}\cos{2y} - \frac{1}{2} = \mathbf{C}$$
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