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Solve \( : x^{2} d y+\left(x y+y^{2}\right) d x=0 \), given that \( y-1 \) when \( x=1 \) |
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Answer» x2dy + (xy + y2)dx = 0 \(\frac{dy}{dx} = \frac{-(xy + y^2)}{x^2}\) = \(-\frac{y}{x}-(\frac{y}{x})^2\) Let y = vu \(\frac{dy}{dx} = v + x \frac{dv}{dx}\) \(v + x \frac{dv}{dx}\) = -v -v2 \(x \frac{dv}{dx}\) = -(v2 + 2v) \(\frac{dv}{v^2 + 2v} = \frac{-dx}{x}\) \(\Rightarrow \int \frac{dv}{(v + 1)^2 - 1} = -\int \frac{dx}{x}\) \(\Rightarrow \frac{1}{2} log\left|\frac{v + 1 - 1}{v + 1 + 1}\right| = -log\,x + log\, c\) \(\Rightarrow\) \(log \left(\frac{v}{v + 2}\right) = 2\,log\,\frac{c}{x}\) \(\Rightarrow\) \(log \left(\frac{y/x}{\frac{y}{x} + 2}\right) = log\,\left(\frac{c}{x}\right)^2\) \(\Rightarrow\) \(\frac{y}{y + 2x} = \frac{c^2}{x^2}\) \(\therefore\) when x = 1, y = 1 \(\therefore\) c2 = \(\frac{1}{1 + 2} = \frac{1}{3}\) \(\therefore\) solution of given differential equation is \(\frac{y}{y + 2x} = \frac{1}{3x^2}\) \(\Rightarrow\) y + 2x = 3x2y \(\Rightarrow\) y + 2x - 3x2y = 0. |
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