Separable Differential Equation Solver

Math Calculus • Differential Equations

Written by STEM Calculators Team Published December 31, 2025 Updated February 24, 2026

1. Separable Differential Equation Solver

Solve separable IVPs of the form \(\;y' = f(x)\,g(y)\;\) with an initial condition \(y(x_0)=y_0\). This tool shows the separation steps and uses RK4 for the plotted solution curve.

Equation input

Input mode: \(f(x)\): \(g(y)\): RHS \(F(x,y)\):

Use x and y. Supports pi, e, sqrt(), sin, cos, exp, etc. Use * for multiplication. In “auto” mode, the tool only succeeds when RHS is literally a product of an x-only part and a y-only part.

Initial condition + evaluation

\(x_0\): \(y_0=y(x_0)\): Evaluate at \(x^\*\):

The “Results” section will include a numerical estimate for \(y(\x\^\*)\) and will show the definite-integral form \(\int_{y_0}^{y}\frac{1}{g(s)}\,ds = \int_{x_0}^{x} f(t)\,dt\).

Graph settings (optional)

x min: x max: y min: y max: RK4 steps: Engine preference:

Drag to pan; mouse wheel to zoom; double-click resets. Wide outputs scroll horizontally on small screens.

Ready

Enter the functions and click “Solve”.

Graph

RK4 solution curve (pan/zoom) — axes/ticks shown

Initial point \((x_0,y_0)\) and the evaluated point at \(x^\*\) are labeled with coordinates.

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Theory — Separable Differential Equations

What “separable” means

A first-order differential equation is separable if it can be written as \[ \frac{dy}{dx} = f(x)\,g(y). \] The key idea is that the \(x\)-dependence and \(y\)-dependence appear as a product.

Examples: \(\;y' = x y\;\) (separable), \(\;y' = x + y\;\) (not separable), \(\;y' = x\sin(y)\;\) (separable).

Separation and integration

Starting from \(\displaystyle \frac{dy}{dx} = f(x)\,g(y)\), move all \(y\)-terms to the left and all \(x\)-terms to the right: \[ \frac{1}{g(y)}\,dy = f(x)\,dx. \]

Integrate both sides: \[ \int \frac{1}{g(y)}\,dy = \int f(x)\,dx + C. \] This is often an implicit equation for \(y\).

Be careful: dividing by \(g(y)\) assumes \(g(y)\neq 0\). If \(g(y)=0\) for some \(y=y_e\), then \(y(x)=y_e\) is an equilibrium (constant) solution.

Using an initial condition (IVP)

If you are given an initial condition \(y(x_0)=y_0\), it is clean to write the solution using definite integrals: \[ \int_{y_0}^{y(x)} \frac{1}{g(s)}\,ds = \int_{x_0}^{x} f(t)\,dt. \] This form automatically “absorbs” the constant \(C\).

If the integrals can be evaluated and you can solve for \(y\), you get an explicit solution \(y=y(x)\). Otherwise, the definite-integral form still defines the solution implicitly.

Worked example: \(y' = x y,\; y(0)=1\)

Here \(f(x)=x\) and \(g(y)=y\). Separate variables: \[ \frac{1}{y}\,dy = x\,dx. \] Integrate: \[ \ln|y| = \frac{x^2}{2} + C. \] Exponentiate: \[ y = C e^{x^2/2}. \] Apply \(y(0)=1\): \(1 = C e^0\Rightarrow C=1\). Therefore \[ y=e^{x^2/2}. \]

Common models: exponential decay and logistic growth

Exponential decay: \[ y'=-k y,\quad k>0. \] Then \(\frac{1}{y}dy=-k\,dx\Rightarrow y=C e^{-kx}\).

Logistic growth: \[ y' = r y\left(1-\frac{y}{K}\right),\quad r,K>0. \] This is separable and has equilibria at \(y=0\) and \(y=K\). Its solutions (for \(0<y_0<K\)) increase toward \(K\).

In many realistic models, explicit closed forms exist, but numeric plotting (direction field + RK4 curve) is still useful.

Frequently Asked Questions

What is a separable differential equation?

A first-order ODE is separable if it can be written as y' = f(x) g(y), where the x-dependence and y-dependence appear as a product. This allows rewriting it as (1/g(y)) dy = f(x) dx and integrating both sides.

How does the solver handle Single RHS mode?

Single RHS mode attempts to factor the right-hand side into an x-only part times a y-only part. It succeeds only when the expression is literally a product of those independent parts.

How is the initial condition y(x0)=y0 used?

After integrating, the constant of integration is determined by substituting x=x0 and y=y0 into the integrated equation. This produces an IVP-specific implicit solution and supports evaluating y at x*.

Why does the graph use RK4 instead of plotting the implicit solution directly?

Many separable solutions are implicit or require solving for y numerically, which can be difficult to plot reliably for all cases. RK4 provides a stable numerical solution curve over the chosen x-interval.