The solution of dy/dx = x + y/x - y is: - Mathematics MCQ Mathematics Mock Test - 8 2026 | ExamTest.live

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The solution of \frac{d y}{d x} = \frac{x + y}{x - y} is:

Solution & Step-by-step Explanation

This is a homogeneous differential equation. Let y = vx, so \frac{d y}{d x} = v + x \frac{d v}{d x} .Substituting: v + x \frac{d v}{d x} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v} x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v ^{2}}{1 - v} = \frac{1 + v ^{2}}{1 - v} Separating variables: \frac{1 - v}{1 + v ^{2}} d v = \frac{d x}{x} Integrating: \int \frac{1}{1 + v ^{2}} d v - \int \frac{v}{1 + v ^{2}} d v = \int \frac{1}{x} d x tan^{- 1} v - \frac{1}{2} lo g (1 + v^{2}) = lo g x + c^{'} Rearranging and substituting v = y / x : lo g x = tan^{- 1} (y / x) - \frac{1}{2} lo g (1 + (y / x)^{2}) + c

Try it yourself before checking the explanation above.

The solution of \frac{d y}{d x} = \frac{x + y}{x - y} is:

lo g x = tan^{- 1} \frac{y}{x} - \frac{1}{2} lo g (1 + (y / x)^{2}) + c

lo g x = tan \frac{y}{x} + \frac{1}{2} lo g (1 + (y / x)^{2}) + c

lo g x = tan^{- 1} \frac{y}{x} + \frac{1}{2} lo g (1 - (y / x)^{2}) + c

lo g x = tan \frac{x}{y} + \frac{1}{2} lo g (1 + (x / y)^{2}) + c

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