RÓWNANIE BERNOULLIEGO
PRZYKŁAD
$$xy^{'} + y = y^{2}\ln x\ / \bullet \frac{1}{x}$$
$$y^{'} + \frac{\mathbf{1}}{\mathbf{x}}y = \frac{\ln\mathbf{x}}{\mathbf{x}} \bullet y^{2}/ \bullet \frac{1}{y^{2}}$$
$$\frac{y^{'}}{y^{2}} + \frac{\mathbf{1}}{\mathbf{x}} \bullet \frac{y}{y^{2}} = \frac{\ln\mathbf{x}}{\mathbf{x}}$$
n = 2
$$y^{- 2}y^{'} + \frac{\mathbf{1}}{\mathbf{x}}y^{- 1} = \frac{\ln\mathbf{x}}{\mathbf{x}}$$
y^−1 = z
−y^−2•y^′ = z^′
$y^{- 2}y^{'} + \frac{\mathbf{1}}{\mathbf{x}}y^{- 1} = \frac{\ln\mathbf{x}}{\mathbf{x}}\mathbf{\ }/ \bullet \left( - 1 \right)\mathbf{\ \rightarrow \ -}\mathbf{y}^{\mathbf{- 2}}\mathbf{y}^{\mathbf{'}} - \frac{1}{x}y^{- 1} = - \frac{\ln x}{x}$ $$z^{'}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}z = \mathbf{-}\frac{\ln\mathbf{x}}{\mathbf{x}}$$
$$\mathbf{z}^{\mathbf{'}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{z = 0}$$
$\frac{\text{dz}}{\text{dx}} = \frac{z}{x}$ $/ \bullet dx/ \bullet \frac{1}{z}$
$$\frac{\text{dz}}{z} = \frac{\text{dx}}{x}$$
$$\int_{}^{}{\frac{1}{z}\text{dz}} = \int_{}^{}{\frac{1}{x}\text{dx}}$$
ln|z| = ln|x| + ln|C|
ln|z| = ln|x•C|
z = Cx
z = C(x)•x
z^′ = C^′(x)x + C(x)•(x)^′
$$\mathbf{C}^{\mathbf{'}}\left( \mathbf{x} \right)\mathbf{x + C}\left( \mathbf{x} \right) \bullet \frac{1}{x} \bullet \mathbf{C}\left( \mathbf{x} \right)\mathbf{x} = - \frac{\ln x}{x}$$
$$C^{'}\left( x \right)x = - \frac{\ln x}{x}/ \bullet \frac{1}{x} \rightarrow \ C^{'}\left( x \right) = - \frac{\ln x}{x^{2}}$$
$$\frac{\text{dC}\left( x \right)}{\text{dx}} = - \frac{\ln x}{x^{2}}/ \bullet dx/ \bullet \int_{}^{}{}$$
$\int_{}^{}{\text{dC}\left( x \right)} = \mathbf{-}\int_{}^{}\frac{\ln\mathbf{x}}{\mathbf{x}^{\mathbf{2}}}\mathbf{\text{dx}}$ $\left| \begin{matrix} u = \ln x/' & v^{'} = x^{- 2} \\ u^{'} = \frac{1}{x} & v = - \frac{1}{x} \\ \end{matrix} \right|$
$$C\left( x \right) = \frac{\ln x}{x} - \int_{}^{}{\frac{1}{x^{2}}\text{dx}} = \frac{\ln x}{x} + \frac{1}{x} + C_{1}$$
$$z = \frac{\ln x}{x} + \frac{1}{x} + C_{1}\text{\ \ \ } \rightarrow \text{\ \ \ }\frac{1}{y} = \frac{\ln x + 1 + xC_{1}}{x}\ \ \rightarrow \text{\ \ }\mathbf{\ \ y =}\frac{\mathbf{x}}{\ln\mathbf{x}\mathbf{+ 1 + x}\mathbf{C}_{\mathbf{1}}}$$