Author Topic: TUT0302 Quiz1  (Read 576 times)

Aiting Zhang

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TUT0302 Quiz1
« on: September 27, 2019, 02:00:17 PM »
$\frac{d y}{d x}=\frac{x^{2}-3 y^{2}}{2 x y}$
$u=\frac{y}{x} \quad y=u \cdot x$
$\frac{d y}{d x}=\frac{d(u \cdot x)}{d x}=\frac{d u}{d x} \cdot x+u \cdot 1$
$\frac{d y}{d x}=\frac{1-\frac{3 y^{2}}{x^{2}}}{\frac{2 x y}{x^2}}=\frac{1-\frac{3 y^{2}}{x^{2}}}{\frac{2 y}{x}}$
$\frac{d y}{d x}=\frac{1-3 u^{2}}{2 u}$
$\frac{d u}{d x} \cdot x+u=\frac{1-3 u^{2}}{2 u}$
$\frac{d u}{d x} \cdot x=\frac{1-5 u^{2}}{2 u}$
$\int \frac{2 u}{1-5 u^{2}} d u=\int \frac{1}{x} d x$
$-\frac{1}{5}\ln| 1-5 u^{2}|=\ln | x |+c$
$e^{-\frac{1}{5} \ln \left|1-5 u^{2}\right|}=e^{\ln |x|+c}$
$e^{\ln \left|1-5 u^{2}\right|^{-\frac{1}{5}}}=e^{\ln |x |} \cdot e^{c}$
$x e^{c}=\left(1-5 u^{2}\right)^{-\frac{1}{5}}$
$x e^{c}=\left(1-5 \frac{y^{2}}{x^{2}}\right)^{-\frac{1}{5}}$