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The general solution of the differential equation $\frac{dy}{dx}+4x\, y^2 =0 $ is : |
$y=\frac{1}{2x^2-C}, $ where C is a constant $y=2x^2-C,$ where C is a constant $y=x^2-C,$ where C is a constant $y=\frac{1}{x^2-C}, $ where C is a constant |
$y=\frac{1}{2x^2-C}, $ where C is a constant |
The correct answer is Option (1) → $y=\frac{1}{2x^2-C}, $ where C is a constant $\frac{dy}{dx}=-4xy^2⇒\int\frac{1}{y^2}dy=\int-4xdx$ $⇒-\frac{1}{y}=-2x^2+C$ So $y=\frac{1}{2x^2-C}$ |
