Given differential equation is "\\frac{dy}{dx}+\\frac{x}{1-x\u00b2}y=x\\sqrt{y}"
"\\implies \\frac{1}{\\sqrt{y}} \\frac{dy}{dx} + \\frac{x}{1-x^2} \\sqrt{y} = x" .
Now, let "\\sqrt{y}=t \\implies \\frac{1}{2\\sqrt{y}} {dy}{dx} = \\frac{dt}{dx}"
So, differential equation becomes: "2\\frac{dt}{dx} + \\frac{x}{1-x^2} t = x \\implies \\frac{dt}{dx} +\\frac{1}{2} \\frac{x}{1-x^2} t = \\frac{x}{2}",
this is linear differential equation of form "\\frac{dt}{dx} + P(x) t = Q(x)".
So, Integrating factor(I.F.) = "e^{{\\int} \\frac{1}{2} \\frac{x}{1-x^2} dx} = e^{\\frac{-1}{4} log(1-x^2)} = (1-x^2)^{\\frac{-1}{4}}"
Hence solution is "t (1-x^2)^{\\frac{-1}{4}} = \\int {\\frac{x}{2} (1-x^2)^{\\frac{-1}{4}}} dx + c"
"\\implies t (1-x^2)^{\\frac{-1}{4}} = (\\frac{1}{-4}) (\\frac{4}{3}) (1-x^2)^{\\frac{3}{4}} + c"
"\\implies \\sqrt{y} (1-x^2)^{\\frac{-1}{4}} = (\\frac{-1}{3}) (1-x^2)^{\\frac{3}{4}} + c"
Now given y(0)=1
"\\implies 1 = \\frac{-1}{3} + c \\implies c = \\frac{4}{3}".
Hence the final solution is "\\sqrt{y} = (\\frac{-1}{3}) (1-x^2) + \\frac{4}{3} (1-x^2)^{\\frac{1}{4}}".
Comments
Leave a comment