Author Topic: TUT0402 Quiz1  (Read 494 times)

Linqian Shen

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TUT0402 Quiz1
« on: September 27, 2019, 02:01:57 PM »
Show that the given equation is homogenous and solve the differential equation:
$$
\frac{d y}{d x}=-\frac{(4 x+3 y)}{(2 x+y)}
$$
since the equation can be written as a ratio of y/x, it is homogenous
$$
\frac{d y}{d x}=-\frac{(4 x+3 y)}{(2 x+y)}=-\frac{\left(4+3 \frac{y}{x}\right)}{\left(2+\frac{y}{x}\right)}
$$
$$
u=\frac{y}{x} \quad y=u x
$$
$$
\frac{d y}{d x}=\frac{d u}{d x} \cdot x+u=-\frac{(4+3 u)}{2+u}
$$
$$
\frac{\partial u}{d x} \cdot x=\frac{-4-3 u-2 u-u^{2}}{2+u}=\frac{-4-5 u-u^{2}}{2+u}
$$
$$
\begin{array}{l}{\frac{2+u}{u^{2}+5 u+4} \partial u=-\frac{1}{x} d x} \\ {\frac{2+u}{(u+1)(u+4)} \quad \partial u=-\frac{1}{x} d x}\end{array}
$$
$$
\begin{array}{l}{\frac{1}{3} \int \frac{1}{u+1} d u+\frac{2}{3} \int \frac{1}{u+4} d u=-\frac{1}{x} d x} \\ {\frac{1}{3} \ln |u+1|+\frac{2}{3} \ln |u+4|=-\ln |x|+C}\end{array}
$$
$$
\frac{1}{3} \ln \left|\frac{y+x}{x}\right|+\frac{2}{3} \ln \left|\frac{y+4 x}{x}\right|=-\ln |x|+C
$$
$$
\frac{1}{3} \ln |y+x|-\frac{1}{3} \ln |x|+\frac{2}{3} \ln |y+4 x|-\frac{2}{3} \ln |x|=-\ln |x|+c
$$
$$
\begin{array}{c}{\ln |y+x|+2 \ln |y+4 x|=36} \\ {\ln |y+x|y+4 x|^{2}=36} \\ {|y+x||y+4 x|^{2}=c}\end{array}
$$