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AP Calculus BC/Unit 7: Differential Equations
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Particular Solution to a Differential Equation

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What is the particular solution to the differential equation dydx=xy2\displaystyle \frac{dy}{dx} = xy^2 satisfying the initial condition y(2)=1\displaystyle y(2) = 1?

A

y=ex22\displaystyle y = e^{\frac{x^2}{2}}

B

y=6−x22\displaystyle y = \frac{6 - x^2}{2}

C

y=ex22−2\displaystyle y = e^{\frac{x^2}{2} - 2}

D

y=26−x2\displaystyle y = \frac{2}{6 - x^2}

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