Question

Show that the coefficients of the given differential equation are homogeneous and solve the given differential equation: $-ydx+(x+√xy)dy=0$

Answer 1

$-ydx+(x+\sqrt{xy})dy=0\\ (x+\sqrt{xy})dy=ydx\\ y'=\frac{y}{x+\sqrt{xy}} \equiv f(x,y)\\ f(tx,ty)=\frac{ty}{tx+\sqrt{tx\cdot ty}}=\\ =\frac{ty}{t(x+\sqrt{x y})}=\frac{y}{x+\sqrt{xy}}=f(x,y)\\ y=x\cdot z(x), y'=x\cdot z'+z\\ xz'+z=\frac{xz}{x+\sqrt{x\cdot xz}}\\ xz'+z=\frac{z}{1+\sqrt{z}}\\ xz'=\frac{z}{1+\sqrt{z}}-z\\ xz'=\frac{-z\cdot\sqrt{z}}{1+\sqrt{z}}\\ \frac{1+\sqrt{z}}{z\cdot\sqrt{z}}dz=-\frac{dx}{x}\\ \int(\frac{1}{z^{\frac{3}{2}}}+\frac{1}{z})dz=-\int\frac{dx}{x}\\ -2 z^{-\frac{1}{2}}+\ln|z|=-\ln|x|-\ln|C|\\ 2 z^{-\frac{1}{2}}-\ln|z|=\ln|x|+\ln|C|\\ \frac{e^{2z^{-\frac{1}{2}}}}{z}=x\cdot C$

return to y

$\frac{e^{2(\frac{y}{x})^{-\frac{1}{2}}}}{\frac{y}{x}}=x\cdot C\\ e^{2(\frac{y}{x})^{-\frac{1}{2}}}=Cy.$