What differential equation formats are supported?

The solver handles first-order separable equations structured as dy/dx = g(x)·h(y).

How are quadratic y-terms like y(1−y) integrated?

The tool uses partial fraction decomposition on 1/h(y) before integrating each logarithmic term.

How do I calculate a particular solution instead of a general solution?

Provide numerical values for both x₀ and y₀ to evaluate the constant C.

What happens if y cannot be algebraically isolated?

The tool outputs the exact implicit equation H(y) = G(x) + C.

Which mathematical expressions can be entered for g(x)?

You can use powers (x^n), linear forms (kx+m)^n, exponentials e^(kx), trigonometric terms, logarithms, and square roots.

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First-Order ODE Solver (Separation of Variables)

Solve separable dy/dx = g(x)·h(y) symbolically: H(y) = G(x) + C, explicit inversion, initial conditions, and partial fractions for quadratics like y(1−y).

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2 examples

dy/dx = 2x·y with y(0) = 1 → y = e^(x²)

Separating gives dy/y = 2x dx; integrating, ln|y| = x² + C, so y = e^(x²+C). The initial condition y(0) = 1 forces C = 0 and the particular solution y = e^(x²).

dy/dx = g(x)·h(y) with g(x) = 2 x, h(y) = y Step 1 — Separate variables: (1/y) dy = 2 x dx Step 2 — Integrate both sides: ln|y| = x^2 + C Step 3 — Solve for y: y = e^(x^2 + C) Initial condition y(0) = 1: C = H(y0) - G(x0) = 0 Particular solution: y = e^(x^2)

Logistic-style dy/dx = x·y(1−y), y(0) = 2 → implicit solution

h(y) = y(1−y) factors into y and (1−y), so 1/h splits by partial fractions into 1/y − 1/(y−1). Integrating gives ln|y| − ln|y−1| = x²/2 + C; H mixes two log terms, so y stays implicit. With y(0) = 2 the constant is C = ln 2 ≈ 0.693147.

dy/dx = g(x)·h(y) with g(x) = x, h(y) = y - y^2 Step 1 — Separate variables (partial fractions): (1/y - 1/(y - 1)) dy = x dx Step 2 — Integrate both sides: ln|y| - ln|y - 1| = 0.5 x^2 + C Step 3 — H(y) mixes irreducible terms: y cannot be isolated in elementary form; the implicit relation above is the solution. Initial condition y(0) = 2: C = H(y0) - G(x0) = 0.693147 Implicit particular solution: ln|y| - ln|y - 1| = 0.5 x^2 + 0.693147

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4 options

ContentPaste or type the main input values.2

g(x) — x-side factortextRequiredx-dependent factor of dy/dx = g(x)·h(y). Terms: constants, x^n, (kx+m)^n, e^(kx), sin(kx+m), cos(kx+m), ln(kx+m), sqrt(kx+m).h(y) — y-side factortextRequiredy-dependent factor. Single terms (y, y^n, (ky+m)^n, e^(ky)) or a degree ≤ 2 polynomial with real roots, or c/D(y).

SettingsAdjust formats, ranges, numbers, and modes.2

x₀ (initial x)numberOptionalOnly used together with y₀ to pin down the constant C.y₀ = y(x₀)numberOptionalInitial value. Leave empty for the general solution; provide with x₀ for the particular solution.

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First-Order ODE Solver (Separation of Variables)

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Examples

1. Separable Exponential Growth Initial Value Problem

2. Logistic Differential Equation with Partial Fractions

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