### Author Topic: TUT5103 Quiz1  (Read 595 times)

#### Yingyingz

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##### TUT5103 Quiz1
« on: September 27, 2019, 02:00:07 PM »
Find the general solution of the equation and prove it is a homogereous equation
$$\because\left(x^{2}+3 y^{2}\right) d x=(2 x y) d y$$
$$\begin{array}{l}{\therefore\text { the equation is homogeneous. }} \\ {\quad y^{\prime}=\frac{1+3 \frac{y^{2}}{x^{2}}}{2 \frac{y}{x}}} \\ {\text { let } u=\frac{y}{x}}\end{array}$$
$$\begin{array}{l}{\therefore y=u x} \\ {\therefore \frac{d y}{d x}=\frac{d(u x)}{d x}=\frac{d u}{d x} x+u} \\ {\therefore \frac{d u}{d x} x=\frac{1+3 u^{2}}{2 u}-\frac{2 u^{2}}{2 u}=\frac{1+u^{2}}{2 u}}\end{array}$$
\begin{aligned} \int \frac{2 u}{1+u^{2}} d u &=\int \frac{1}{x} d x \\ \ln \left|1+u^{2}\right| &=\ln |x|+C \\ e^{\ln \left|1+u^{2}\right|} &=\ln |x|+c \\ e^{\ln \left|1+u^{2}\right|} &=e^{\ln |x|+c} \\ e^{\ln | | t+u^{2} |} &=e^{\ln |x|} \cdot e^{c} \\ 1+u^{2} &=c x \\\left(\frac{y}{x}\right)^{2} &=c x-1 \\ \frac{y^{2}}{x^{2}} &=c x-1 \\ y^{2} &=c x^{3}-x^{2} \end{aligned}