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When solving the common form of the special Ricatti

$$y' = q(x) + p(x)y + r(x)y^2$$

The first step is to assume a solution of:

$$y(x) = y_1(x) + z(x)$$ Which reduces the Differential Equation into a Bernoulli Equation:

$$z(x)' = p(x)z(x) +r(x)[2y_1(x) z(x) + z(x)^2]$$

However, doesn't this imply that the differential eqution has two linearly independant solutions? Shouldn't a Differential Equation of order one contain only one solution?

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This is not a linear equation, so you cannot expect a family of solutions depending linearly on parameters. In particular, the result you are mentioning (a linear differential equation has a general integral of dimension one) does not hold. See comments here. By all means, I cannot understand what makes you think that the second equation has two linearly independent solutions. – Giuseppe Negro Jan 9 '13 at 0:27
Oh, and one last thing: the name is spelled Riccati, not Ricatti. Can you edit, please? – Giuseppe Negro Jan 9 '13 at 0:29

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