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Question
Find the general solution of the differential equation: (x3 + y3)dy = x2ydx
Solution
Given differential equation is (x3 + y3)dy = x2ydx
∴ `(dx)/(dy) = (x^3 + y^3)/(x^2y)` ...(i)
Put x = vy
⇒ `(dx)/(dy) = v + y (dv)/(dy)`
From equation, we have
`v + y (dv)/(dy) = ((vy)^3 + y^3)/((vy)^2y)`
`v + y (dv)/(dy) = (v^3y^3 + y^3)/(v^2y^3)`
`v + y (dv)/(dy) = (v^3 + 1)/v^2`
`y (dv)/(dy) = (v^3 + 1)/v^2 - v`
`y (dv)/(dy) = 1/v^2`
`v^2dv = (dy)/y` ...(Variable seponation method)
Integrating both sides, we get
`int v^2dv = int (dy)/y`
`v^3/3` = log + C
Putting v = `x/y`, we get
`x^3/(3y^3)` = log + C
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