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Find the general solution of the following differential equation:\(\frac{dy}{dx}\) = (1 + x)(1 + y2) |
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Answer» \(\frac{1}{1+y^2}dy\) = (1 + x)dx Integrating on both sides \(\int\frac{1}{1+y^2}dy\) = \(\int(1+x)dx\) ⇒ tan-1y = x + \(\frac{x^2}2\) + c |
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