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Solve the differential equation (x-y)dy-(x+y)dx=0

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Solve the differential equation $(x - y) d y - (x + y) d x = 0$

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Solve the differential equation

(x - y) d y - (x + y) d x = 0

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Text solutionVerified

(x - y) d y - (x + y) d x = 0

(x + y) d x = (x - y) d y

\frac{d y}{d x} = \frac{x + y}{x - y}

∴ \frac{d y}{d x} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}}

Let,

\frac{y}{x} = V, y = v - x

\frac{d y}{d x} = v + x . \frac{d v}{d x}

v + x . \frac{d v}{d x} = \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}

∴ \frac{1 - v}{1 + v ^{2}} d v = \frac{1}{x} . d x

Integrating both side

∴ \int \frac{1}{1 + v ^{2}} d v - \frac{1}{2} \int \frac{2 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 + lo g c

∴ tan^{- 1} \frac{y}{x} - \frac{1}{2} lo g ∣1 + \frac{y ^{2}}{x ^{2}} ∣ = lo g x + lo g c

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

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

∴ e^{t a n \frac{y}{x}} = \frac{x . c . ( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}{x} = (x 62 + y^{2})^{\frac{1}{2}} . c

∴ e^{t a n \frac{y}{x}} = (x^{2} + y^{2})^{\frac{1}{2}} . c

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Question Text	Solve the differential equation $(x - y) d y - (x + y) d x = 0$
Answer Type	Text solution:1
Upvotes	150