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Question
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
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Solution
y dx + (x - y2 ) dy = 0
∴ y dx = (y2 - x) dy
∴ `dx/dy = (y^2 - x) /y= y - x/y `
∴ `dx/dy + x/y = y`
The given equation is of the form
`dx/dy + Px = Q`
where, P = `1/y` and Q = y
∴ I.F. = `e int^ (pdy) = e int ^(1/ydy) = e ^(log |y|)= y`
∴ Solution of the given equation is
`x (I.F.) =int Q (I.F.) dy + c_1`
∴ `xy = int y(y) dy = y^3/3 + c_1`
∴ 3xy = y3 + c …[3c1 = c]
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