Multiple choice

The solution for the differential equation $\frac{dy}{dx} = x^2y$ with condition that y = 1 at x = 0 is

  1. $y = e^{\frac{1}{2x}}$
  2. In $(y) = \frac{x^3}{3} + 4$
  3. In $(y) = \frac{x^2}{2}$
  4. $y = e^{\frac{x^3}{3}}$
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D Correct answer
Explanation

Separating variables gives dy/y = x^2 dx. Integrating both sides results in ln(y) = x^3/3 + C. Using the condition y=1 at x=0, we find ln(1) = 0 + C, so C = 0. Thus, ln(y) = x^3/3, which implies y = e^(x^3/3).

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