Question

\left(x-y\right)dx+\left(3x+y\right)dy=0

Answer 1

Let us solve the differential equation $\left(x-y\right)dx+\left(3x+y\right)dy=0.$ For this let us use the transformation $y=ux.$ Then $dy=udx+xdu,$ and we get the equation $\left(x-ux\right)dx+\left(3x+ux\right)(udx+xdu)=0.$ Since $x=0$ is not a solution, let us divide both parts by $x.$ Then we get $\left(1-u\right)dx+\left(3+u\right)(udx+xdu)=0,$ which is equivalent to $\left(1+2u+u^2\right)dx+\left(3+u\right)xdu=0.$ It follows that $\frac{dx}x+\frac{\left(3+u\right)du}{(1+u)^2}=0,$ and hence $\int\frac{dx}x+\int \frac{\left(1+u+2\right)du}{(1+u)^2}=C.$ We conclude that $\ln|x|+\int \frac{du}{1+u}+2\int \frac{du}{(1+u)^2}=C,$ and hence $\ln|x|+\ln|1+u|-2\frac{1}{1+u}=C.$

We conclude that $\ln|x|+\ln|1+\frac{y}x|-\frac{2x}{x+y}=C$ is the general solution of the equation $\left(x-y\right)dx+\left(3x+y\right)dy=0.$