Express 2x+1 / x(x+1) in partial fractions and hence find the general solution of the differential equation x(x+1)dy/dx = y(2x+1) expressing y explicitly in terms of x.
It follows that "\\frac{2x+1}{x(x+1)}=\\frac{1}{x}+\\frac{1}{x+1}" is the expression in partial fractions. Let us find the general solution of the differential equation "x(x+1)\\frac{dy}{dx} = y(2x+1)" expressing "y" explicitly in terms of "x". We get that "\\frac{dy}{y} = \\frac{2x+1}{x(x+1)}dx," and hence "\\int\\frac{dy}{y} = \\int\\frac{2x+1}{x(x+1)}dx=\\int(\\frac{1}{x}+\\frac{1}{x+1})dx." We conclude that "\\ln|y|=\\ln|x|+\\ln|x+1|+\\ln|C|=\\ln|x(x+1)C|." Therefore, "y=Cx(x+1)" is the general solution of the differential equation "x(x+1)\\frac{dy}{dx} = y(2x+1)".
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