Równanie jednorodne, podstawienie z t
: 16 cze 2012, o 16:45
Proszę o sprawdzenie:
\(\displaystyle{ y'= \frac{2x^{2}-y^{2}}{xy} \qquad x,y \neq 0 \\
t= \frac{y}{x} \\
t+xt'= \frac{2-t^{2}}{t} \qquad t \neq 0 \\
x \frac{dt}{dx} = \frac{2-2t^{2}}{t} \\
\int \frac{tdt}{2-2t^{2}} = \int \frac{dx}{x} \qquad t \neq 1\; \qquad t \neq -1 \\
- \frac{1}{4} \ln |2-2t^{2}|=\ln|x|+\ln C_1 \\
(|2-2t^{2}|)^{ -\frac{1}{4} } = C_1\cdotx \qquad C_1>0 \\
(2-2t^{2})^{-\frac{1}{4}}=C_2x \qquad C_2 \neq 0 \\
2-2t^{2}=C_3x^{-4} \qquad C_3 \neq 0 \\
t^{2}=1-C_4x^{-4} \qquad C_4=\frac{1}{2} \cdot C_3 \\
t = \sqrt{1-C_4x^{-4}} \qquad t= -\sqrt{1-C_4x^{-4}}\\
y = x\cdot \sqrt{1-C_4x^{-4}}\ lub \ y = -x\cdot\sqrt{1-C_4x^{-4}}}\)
\(\displaystyle{ y'= \frac{2x^{2}-y^{2}}{xy} \qquad x,y \neq 0 \\
t= \frac{y}{x} \\
t+xt'= \frac{2-t^{2}}{t} \qquad t \neq 0 \\
x \frac{dt}{dx} = \frac{2-2t^{2}}{t} \\
\int \frac{tdt}{2-2t^{2}} = \int \frac{dx}{x} \qquad t \neq 1\; \qquad t \neq -1 \\
- \frac{1}{4} \ln |2-2t^{2}|=\ln|x|+\ln C_1 \\
(|2-2t^{2}|)^{ -\frac{1}{4} } = C_1\cdotx \qquad C_1>0 \\
(2-2t^{2})^{-\frac{1}{4}}=C_2x \qquad C_2 \neq 0 \\
2-2t^{2}=C_3x^{-4} \qquad C_3 \neq 0 \\
t^{2}=1-C_4x^{-4} \qquad C_4=\frac{1}{2} \cdot C_3 \\
t = \sqrt{1-C_4x^{-4}} \qquad t= -\sqrt{1-C_4x^{-4}}\\
y = x\cdot \sqrt{1-C_4x^{-4}}\ lub \ y = -x\cdot\sqrt{1-C_4x^{-4}}}\)