Answer to Question #300201 in Differential Equations for Joshua

Question #300201

Show that the coefficients of the given differential equation are homogeneous and solve the given differential equation: "-ydx+(x+\u221axy)dy=0"

Expert's answer

"-ydx+(x+\\sqrt{xy})dy=0\\\\\n(x+\\sqrt{xy})dy=ydx\\\\\ny'=\\frac{y}{x+\\sqrt{xy}} \\equiv f(x,y)\\\\\nf(tx,ty)=\\frac{ty}{tx+\\sqrt{tx\\cdot ty}}=\\\\\n=\\frac{ty}{t(x+\\sqrt{x y})}=\\frac{y}{x+\\sqrt{xy}}=f(x,y)\\\\\n y=x\\cdot z(x), y'=x\\cdot z'+z\\\\\nxz'+z=\\frac{xz}{x+\\sqrt{x\\cdot xz}}\\\\\nxz'+z=\\frac{z}{1+\\sqrt{z}}\\\\\nxz'=\\frac{z}{1+\\sqrt{z}}-z\\\\\nxz'=\\frac{-z\\cdot\\sqrt{z}}{1+\\sqrt{z}}\\\\\n\\frac{1+\\sqrt{z}}{z\\cdot\\sqrt{z}}dz=-\\frac{dx}{x}\\\\\n\\int(\\frac{1}{z^{\\frac{3}{2}}}+\\frac{1}{z})dz=-\\int\\frac{dx}{x}\\\\\n-2 z^{-\\frac{1}{2}}+\\ln|z|=-\\ln|x|-\\ln|C|\\\\\n2 z^{-\\frac{1}{2}}-\\ln|z|=\\ln|x|+\\ln|C|\\\\\n\\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\\\\\ne^{2(\\frac{y}{x})^{-\\frac{1}{2}}}=Cy."