Variation of Parameter Tool

Math Calculus • Differential Equations

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

7. Variation of Parameters Tool

Find a particular solution of $y'' + p(x)\,y' + q(x)\,y = r(x)$ using $$y_p=u_1y_1+u_2y_2$$ with a homogeneous basis $\{y_1,y_2\}$ . Computes the Wronskian and the formulas $u_1'=-\frac{y_2 r}{W}$ , $u_2'=\frac{y_1 r}{W}$ .

Input supports: pi, e, sqrt(), sin, cos, tan, exp, log. Use * for multiplication. Variable is x.

$p(x)$: $q(x)$: $r(x)$:

Homogeneous basis $\{y_1(x),y_2(x)\}$

$y_1(x)$: $y_2(x)$:

The tool checks the Wronskian $W=y_1y_2' - y_1'y_2$. If $W\approx 0$ on your interval, the basis is not valid there.

Optional initial conditions (to get a full solution)

Solve $C_1,C_2$ from IC: $x_0$: $y(x_0)=y_0$: $y'(x_0)=v_0$:

Graph marks $(x_0,y(x_0))$ and $(x^\*,y(x^\*))$ with coordinate labels.

Verification + Graph settings (optional)

$x_{\min}$: $x_{\max}$: Samples: Mark on graph $x^\*$: Check basis in ODE: Engine preference:

Drag to pan, mouse wheel to zoom, double-click to reset view. Outputs scroll horizontally on small screens.

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Steps

Enter $p,q,r$ and a basis $\{y_1,y_2\}$, then click Solve.

Graph

Plot of $y_p(x)$ (and $y(x)$ if IC enabled) — pan/zoom

Hover to probe $(x,y)$.

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Theory

Variation of Parameters (2nd-order linear ODE)

Consider the non-homogeneous linear ODE \[ y'' + p(x)\,y' + q(x)\,y = r(x). \] Let $y_1(x)$ and $y_2(x)$ be a fundamental set of solutions of the homogeneous equation \[ y'' + p(x)\,y' + q(x)\,y = 0. \] Then every solution of the homogeneous equation is \[ y_h(x)=C_1y_1(x)+C_2y_2(x). \]

Wronskian

The Wronskian of $y_1,y_2$ is \[ W(x)= \begin{vmatrix} y_1 & y_2\\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_1'y_2. \] On intervals where $W(x)\neq 0$, the pair $\{y_1,y_2\}$ is linearly independent and forms a valid basis.

Particular solution ansatz

Variation of parameters seeks a particular solution of the form \[ y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x), \] where $u_1,u_2$ are unknown functions. A standard derivation imposes the auxiliary condition \[ u_1'(x)y_1(x)+u_2'(x)y_2(x)=0, \] which simplifies the algebra.

Formulas for $u_1',u_2'$

With the above condition, one obtains the classic system \[ \begin{aligned} u_1'(x)y_1(x) + u_2'(x)y_2(x) &= 0,\\ u_1'(x)y_1'(x) + u_2'(x)y_2'(x) &= r(x). \end{aligned} \] Solving it gives \[ u_1'(x) = -\frac{y_2(x)\,r(x)}{W(x)}, \qquad u_2'(x) = \frac{y_1(x)\,r(x)}{W(x)}. \]

Integrating to obtain $u_1,u_2$

Typically, \[ u_1(x)=\int u_1'(x)\,dx,\qquad u_2(x)=\int u_2'(x)\,dx, \] and then \[ y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x). \] Any constants from these integrals can be absorbed into $C_1,C_2$ in the homogeneous part.

Full solution

The general solution of the original ODE is \[ y(x)=y_h(x)+y_p(x)=C_1y_1(x)+C_2y_2(x)+y_p(x). \] If initial conditions $y(x_0)=y_0$, $y'(x_0)=v_0$ are given, substitute them to solve for $C_1,C_2$.

Tip: If $r(x)$ is a “nice” forcing term and the coefficients are constant, you may compare with the method of undetermined coefficients. Variation of parameters works more generally but often requires integration.

Frequently Asked Questions

What does the variation of parameters tool solve?

It targets non-homogeneous second-order linear ODEs of the form y'' + p(x) y' + q(x) y = r(x). With a provided homogeneous basis y1 and y2, it constructs a particular solution yp and can also produce the full solution y = C1 y1 + C2 y2 + yp.

How do I choose y1(x) and y2(x) for variation of parameters?

y1 and y2 must be linearly independent solutions of the homogeneous equation y'' + p(x) y' + q(x) y = 0. A quick indicator is a nonzero Wronskian W(x) on the interval of interest.

Why does a near-zero Wronskian mean the basis is not valid?

If W(x) = y1 y2' - y1' y2 is approximately zero on an interval, y1 and y2 behave as dependent there and do not form a fundamental set. The formulas u1' = -y2 r / W and u2' = y1 r / W then become unstable or undefined.

How is the particular solution yp(x) computed in this tool?

The tool uses yp = u1 y1 + u2 y2 with u1' = -y2 r / W and u2' = y1 r / W, where W is the Wronskian of y1 and y2. It integrates u1' and u2' to obtain u1 and u2 and then forms yp.

Can the tool use initial conditions to find C1 and C2?

Yes, when initial conditions are enabled it solves for C1 and C2 from y(x0) = y0 and y'(x0) = v0. The resulting full solution is then plotted and can be inspected alongside the step-by-step work.

Steps

Graph

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Frequently Asked Questions

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