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प्रश्न
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
पर्याय
\[y^2 = \exp\left( x + \frac{x^2}{2} \right) - 1\]
\[y^2 = 1 + C \exp\left( x + \frac{x^2}{2} \right)\]
y = tan (C + x + x2)
\[y = \tan\left( x + \frac{x^2}{2} \right)\]
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