Solution:
To solve differential equation: "\\frac{dy}{dx} + \\frac y x=\\frac{y^2}{x^2}"
"\\frac{dy}{dx} + \\frac y x=\\frac{y^2}{x^2} ...(i)"
Put "y=vx ...(ii)"
On differentiating w.r.t "x"
"\\frac{dy}{dx}=v+\\frac{xdv}{dx} ...(iii)"
Put (ii) and (iii) in (i), we get,
"v+\\frac{xdv}{dx}+\\frac{vx}{x}=\\frac{v^2x^2}{x^2}\n\\\\ \\Rightarrow v+\\frac{xdv}{dx}+v=v^2\n\\\\ \\Rightarrow \\frac{xdv}{dx}=v^2-2v\n\\\\ \\Rightarrow \\frac{dv}{v^2-2v}=\\frac{dx}{x}\n\\\\ \\Rightarrow \\frac{dv}{(v-1)^2-1^2}=\\frac{dx}{x}"
On integrating both sides,
"\\frac 1 2 \\ln |\\frac{(v-1)-1}{(v-1)+1}|=\\ln x+C\n\\\\ \\Rightarrow \\frac 1 2 \\ln |\\frac{v-2}{v}|=\\ln x+C\n\\\\ \\Rightarrow \\ln |\\frac{v-2}{v}|=2\\ln x+2C"
"\\\\ \\Rightarrow \\ln |\\frac{\\frac y x-2}{\\frac y x}|=2\\ln x+2C" [Using (ii)]
"\\\\ \\Rightarrow \\ln |\\frac{y-2x}{y}|=\\ln x^2+2C \n\\\\ \\Rightarrow \\ln |\\frac{y-2x}{y}|-\\ln x^2=2C \n\\\\ \\Rightarrow \\ln |\\frac{y-2x}{yx^2}|=2C"
"\\\\ \\Rightarrow\\frac{y-2x}{yx^2}=e^{2C}" [When "\\ln y=x", then "y=e^x"]
"\\\\ \\Rightarrow\\frac{y-2x}{yx^2}=k" [where "k=e^{2C}", constant]
"\\Rightarrow y-2x=kyx^2\n\\\\ \\Rightarrow y-kyx^2=2x\n\\\\ \\Rightarrow y(1-kx^2)=2x\n\\\\ \\Rightarrow y=\\frac{2x}{1-kx^2}"
Answer:
The solution of given differential equation is "y=\\frac{2x}{1-kx^2}" , where "k" is a constant.
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