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Question
For the differential equation, find the general solution:
`(x + 3y^2) dy/dx = y(y > 0)`
Solution
`(x + 3y^2) dy/dx = y`
or `y dx/dy = x + 3y^2`
`dx/dy - x/y = 3y`
A vertical differential equation of the form `dx/dy + Px = Q.`
Here `P = - 1/y, Q = 3y`
∴ `I.F. = e^(int P dx) = e^(- int 1/y dy) = e^(- log y) = 1/y`
Hence, the general solution of the differential equation
⇒ `x × I.F. = int Q xx I.F. dx + C`
⇒ `x xx 1/y = int 1/y (3y) dy + C`
⇒ `x/y = 3 int 1 dy + C`
⇒ `x/y = 3y + C`
⇒ x = 3y2 + Cy
Which is the required solution.
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