Linear First Order Differential Equation Tool

Math Calculus • Differential Equations

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

2. Linear First Order Differential Equation Tool

Solve linear first-order ODEs: \(\displaystyle y' + P(x)\,y = Q(x)\) using the integrating factor \(\mu(x)=e^{\int P(x)\,dx}\), with an optional initial condition \(y(x_0)=y_0\).

Equation

\(P(x)\): \(Q(x)\):

Input supports pi, e, sqrt( ), sin, cos, exp, etc. Use * for multiplication.

Initial condition (IVP) + probe

\(x_0\): \(y_0=y(x_0)\): Probe \(x^\*\):

The plot shows the RK4 numerical solution through \((x_0,y_0)\). The marker at \(x^\*\) reports the approximated \(y(x^\*)\).

Graph settings (optional)

x min: x max: y min: y max: RK4 steps: Solution family: Family \(y_0\) list:

“Solution family” plots multiple IVPs \(y(x_0)=y_0\) on the same axes (thin colored curves), plus your main IVP (thick curve).

Engine preference:

Ready

Enter \(P(x)\), \(Q(x)\), and click “Solve”.

Graph

IVP solution (pan/zoom) — axes/ticks shown

Drag to pan; mouse wheel to zoom; double-click resets. Axes are \(x\) (horizontal) and \(y\) (vertical) in your input units.

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Theory — Linear First-Order Differential Equations

Standard form

A linear first-order ODE has the form \[ y' + P(x)\,y = Q(x), \] where \(P(x)\) and \(Q(x)\) are known functions of \(x\).

“Linear” means \(y\) and \(y'\) appear only to the first power and are not multiplied together (no \(y^2\), no \(y\,y'\), etc.).

Integrating factor method

Define the integrating factor \[ \mu(x) = e^{\int P(x)\,dx}. \] Multiply the ODE by \(\mu(x)\): \[ \mu y' + \mu P y = \mu Q. \] The left-hand side becomes a product derivative: \[ \frac{d}{dx}\big(\mu(x)\,y(x)\big) = \mu(x)\,Q(x). \] Integrate: \[ \mu(x)\,y(x) = \int \mu(x)\,Q(x)\,dx + C, \] so \[ y(x) = \frac{\int \mu(x)\,Q(x)\,dx + C}{\mu(x)}. \]

Using an initial condition \(y(x_0)=y_0\)

A clean way to incorporate the IVP is to use definite integrals: \[ \mu(x)\,y(x) - \mu(x_0)\,y_0 = \int_{x_0}^{x} \mu(t)\,Q(t)\,dt. \] Therefore, \[ y(x)=\frac{\mu(x_0)\,y_0+\int_{x_0}^{x}\mu(t)\,Q(t)\,dt}{\mu(x)}. \]

Worked example: \(y' + y = e^x,\; y(0)=0\)

Here \(P(x)=1\) and \(Q(x)=e^x\). Then \[ \mu(x)=e^{\int 1\,dx}=e^x. \] Multiply the ODE by \(e^x\): \[ e^x y' + e^x y = e^{2x}. \] Recognize the left side: \[ \frac{d}{dx}(e^x y) = e^{2x}. \] Integrate: \[ e^x y = \frac{e^{2x}}{2} + C. \] Solve for \(y\): \[ y=\frac{1}{2}e^x + C e^{-x}. \] Apply \(y(0)=0\): \[ 0=\frac{1}{2}+C \Rightarrow C=-\frac{1}{2}. \] Therefore \[ y=\frac{1}{2}\big(e^x-e^{-x}\big) \quad\text{or}\quad y=\frac{1}{2}(1-e^{-x})\;\text{if }Q(x)=1\text{ instead of }e^x. \]

Note: For the sample in this chapter, the correct final form depends on the chosen \(Q(x)\). The calculator uses your exact \(P(x)\) and \(Q(x)\) inputs.

Homogeneous + particular viewpoint

The solution can be viewed as \[ y(x)=y_h(x)+y_p(x), \] where \(y_h\) solves the homogeneous equation \(y'+P(x)y=0\), and \(y_p\) is any one particular solution to \(y'+P(x)y=Q(x)\).

The integrating factor method automatically produces the full solution (homogeneous + particular) in one pass.

University extension: Bernoulli reduction

A Bernoulli equation \[ y' + P(x)y = Q(x)y^n \] becomes linear after the substitution \(v=y^{1-n}\) (for \(n\neq 1\)). This tool focuses on the linear case \(n=0\), but the integrating-factor idea reappears after the substitution.

Frequently Asked Questions

What is a linear first-order differential equation?

It is an equation that can be written as y' + P(x) y = Q(x), where P and Q depend only on x. The unknown function y appears only to the first power and is not multiplied by y'.

How does the integrating factor method solve y' + P(x) y = Q(x)?

The method uses mu(x)=e^(int P(x) dx) to turn the left side into a product derivative: d/dx(mu(x) y(x)) = mu(x) Q(x). Integrating then gives mu(x) y(x) = int mu(x) Q(x) dx + C.

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

After integrating, the constant C is determined by substituting x=x0 and y=y0 into the integrated expression. The tool can also express the result using a definite-integral form from x0 to x.

What does the Solution family option do?

It plots multiple solution curves for the same P(x) and Q(x) using different initial values y(x0)=y0 provided in the y0 list. Your main IVP remains highlighted while the family curves are shown as additional overlays.

What does changing the RK4 steps affect?

RK4 steps control the numerical resolution of the plotted solution curve. More steps typically give a smoother and more accurate plot over the chosen x-range.