Question

$\frac{dy}{dx}+\frac{x^2+3y^2}{3x^2+y^2}=0$

Answer 1

$\begin{array}{c}\frac{dy}{dx}+\frac{x^2+3y^2}{3x^2+y^2}=0\\ \Rightarrow \frac{dy}{dx}=-\frac{x^2+3y^2}{3x^2+y^2}\\ puttingy=vx\\ \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \Rightarrow v+x\frac{dv}{dx}=-\frac{x^2+3v^2{x}^2}{3x^2+v^2{x}^2}\\ \Rightarrow v+x\frac{dv}{dx}=-\left(\frac{1+3v^2}{3+v^2}\right)\\ \Rightarrow x\frac{dv}{dx}=\frac{-1-3v^2}{3+v^2}-v\\ \Rightarrow x\frac{dv}{dx}=\frac{-1-3v^2-3v-v^3}{3+v^2}\\ \Rightarrow \int \frac{v^2+3}{v^3+3v^2+3v+1}dv=\int -\frac{dx}{x}\\ \Rightarrow \int \frac{v^2+3}{\left(v+1\right)\left(v^2+2v+1\right)}dv=\int -\frac{dx}{x}\\ \Rightarrow \int \frac{v^2+3}{\left(v+1\right){\left(v+1\right)}^2}dv=\int -\frac{dx}{x}\\ \Rightarrow \int \frac{v^2+3}{{\left(v+1\right)}^3}dv=\int -\frac{dx}{x}\\ puttingv+1=t\\ dv=dt\\ \Rightarrow \int \frac{{\left(t-1\right)}^2+3}{t^3}dt=-\log \left|x\right|\\ \Rightarrow \int \frac{t^2+1-2t+3}{t^3}dt=\log \left|x\right|\\ \Rightarrow \int \frac{t^2-2t+4}{t^3}dt=\log \left|x\right|\\ \Rightarrow \int \left(\frac{1}{t}-\frac{2}{t^2}+\frac{4}{t^3}\right)dt=\log \left|x\right|\\ \Rightarrow \log t+2t^{-1}+4\times \frac{-2}{t^2}=\log \left|x\right|+c\\ \Rightarrow \log t\left|v+1\right|+\frac{2}{v+1}-\frac{8}{{\left(v+1\right)}^2}=\log \left|x\right|+c\\ \Rightarrow \log t\left|x+y\right|-\log \left|x\right|+\frac{2x}{x+y}-\frac{8x^2}{{\left(x+y\right)}^2}=\log \left|x\right|+c\\ \Rightarrow \log t\left|x+y\right|+\frac{2x}{x+y}-\frac{8x^2}{{\left(x+y\right)}^2}=2\log \left|x\right|+c\end{array}$