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प्रश्न
Solve the following differential equation: (3xy + y2) dx + (x2 + xy) dy = 0
उत्तर
(3xy + y2) dx + (x2 + xy) dy = 0
`(dy)/(dx) = - ((3xy + y^2)/(x^2 + xy))`
Put y = vx ⇒ `(dy)/(dx) = v + x (dv)/(dx)`
`v + x (dv)/(dx) = - ((3x . vx + v^2 x^2)/(x^2 + x . vx))`
` x (dv)/(dx) = (-3v - v^2)/(1 + v) - v`
` x (dv)/(dx) = (-3v - v^2 - v - v^2)/(1 + v) `
` x (dv)/(dx) = (-2v^2 - 4v)/(1 + v)`
`(1 + v)/(2v^2 + 4v) "dv" = -(1)/(x) "dx"`
` int_ (1 + v)/(2v^2 + 4v) "dv" = int_ -(1)/(x) "dx"`
` 1/4 int_ (2 + 2v)/(2v + v^2) "dv" = - int_ (1)/(x) "dx"`
`1/4 log | v^2 + 2v| = - log | x | + c`
`1/4 log ((y^2)/(x^2) + 2 (y)/(x)) . x = c `
`log ((y^2)/(x) + 2y) = 4c`
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