Math · Calculus II · Worked example

Find an implicit solution of dy/dx = −x/y

Solve dy/dx = −x/y with y(0) = 4, and give the interval on which the solution is valid.

Separate

Multiply both sides by y dx: y dy = −x dx.

Integrate

y²/2 = −x²/2 + C. At (0, 4), C = x²/2 + y²/2 = 8.

\frac{0^{2}}{2} + \frac{4^{2}}{2} = 8

Write the implicit solution

Multiplying by 2 gives x² + y² = 16, so the solution curve lies on the circle of radius 4. The condition y(0) = 4 picks the upper half: y = √(16 − x²).

Find the interval of validity

The equation divides by y, so y ≠ 0: the solution is valid for −4 < x < 4. At x = ±4 the curve reaches the axis, and the slope −x/y is undefined there.

Check

Differentiating √(16 − x²) gives −x/√(16 − x²), which is −x/y, and the curve passes through (0, 4).

\begin{array}{l} y (x) = \sqrt{16 - x^{2}} \\ y (0) = 4 \end{array}

Result

x² + y² = 16, with the explicit solution y = √(16 − x²) valid on −4 < x < 4.

Your turn

Solve dy/dx = x/y with y(0) = 3.

Show the answer and explanation

y² − x² = 9, so y = √(x² + 9) for every x.

y dy = x dx gives y²/2 = x²/2 + C, and y(0) = 3 gives C = 9/2. Since x² + 9 is never 0, y = √(x² + 9) is valid for all x.

\begin{array}{l} y (x) = \sqrt{x^{2} + 9} \\ y (0) = 3 \end{array}

Keep exploring

Derivative & antiderivative checks verifies that √(16 − x²) has derivative −x/√(16 − x²) on −4 < x < 4.

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OpenStax Calculus Volume 2 — Separable equations

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