RÓWNANIE RÓŻNICZKOWE LINIOWE – METODA UZMIENNIANIA STAŁEJ
PRZYKŁAD
y^′ + 2xy = 0
$$\frac{\text{dy}}{\text{dx}} = - 2xy\ \ / \bullet \frac{\text{dx}}{y}$$
$$\frac{\text{dy}}{y} = - 2xdx\ \ / \bullet \int_{}^{}{}$$
$$\int_{}^{}\frac{1}{y}dy = - 2\int_{}^{}\text{xdx}$$
$$\ln\left| y \right| = - 2 \bullet \frac{1}{2}x^{2} + C$$
y = e^−x2 + C
y = e^−x2 • e^C
CORLJ
y^′+2xy = xe^−x2
CORLN
y^′ = C^′(x) • e^−x2 + C(x) • e^−x2 • (−2x)
$$C^{'}\left( x \right)e^{- x^{2}} - 2xe^{- x^{2}}C\left( x \right) + \mathbf{2}\mathbf{x}C\left( x \right)e^{{- x}^{2}} = \mathbf{x}\mathbf{e}^{\mathbf{- x}^{\mathbf{2}}}\ / \bullet \frac{1}{e^{{- x}^{2}}}$$
C^′(x) − 2xC(x) + 2xC(x) = x
C^′(x) = x
$$\frac{\text{dC}\left( x \right)}{\text{dx}} = x\ / \bullet dx$$
dC(x) = xdx /∫
$$C\left( x \right) = \frac{1}{2}x^{2} + C_{1}$$
$$y = \left( \frac{1}{2}x^{2} + C_{1} \right)e^{{- x}^{2}}$$
$$y = \frac{1}{2}x^{2}e^{{- x}^{2}} + C_{1}e^{{- x}^{2}}$$