If dy/dx = x/y and y(0) = 2, what is y(x)?

Question

Accepted Answer

y = √(x² + 4). The correct answer is y = √(x² + 4). Separating variables: y dy = x dx. Integrating: y²/2 = x²/2 + C. Using y(0) = 2: 4/2 = 0 + C, so C = 2. Therefore y²/2 = x²/2 + 2, giving y² = x² + 4, so y = √(x² + 4) (positive root since y(0) = 2 > 0).

The expression y = x²/2 + 2 does not properly separate variables; it treats the equation as if y dy = x dx integrates to y = x²/2 + C, ignoring the y² that arises on the left side. The expression y = √(2x + 4) has x inside the square root linearly, but integrating x dx gives x²/2, not x. The expression y = x + 2 is a linear function that doesn't satisfy the original differential equation; substituting back gives dy/dx = 1 but x/y = x/(x+2) ≠ 1.

Initial value problems combining separation of variables with an initial condition are standard on the CLEP exam. Students should practice the full process: separate, integrate both sides, apply the initial condition to find C, and solve for y. Always verify the solution satisfies both the equation and the initial condition.

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