Second Order Homogenous Differential Equation Solver

Math Calculus • Differential Equations

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

4. Second Order Homogenous Differential Equation Solver

Solve constant-coefficient homogeneous ODEs of the form \(\;y'' + p\,y' + q\,y = 0\;\) using the characteristic equation. The tool detects distinct/repeated/complex roots, builds the general solution, and (optionally) determines constants from initial conditions.

Equation

\(p\): \(q\):

Supports numeric expressions like pi, e, sqrt(2). The characteristic equation is \(r^2 + p r + q = 0\).

Constants (choose one method)

Determine \(A,B\) by: \(x_0\): \(y_0=y(x_0)\): \(v_0=y'(x_0)\): \(A\): \(B\):

If you choose “Initial conditions”, the tool solves a 2×2 linear system for \(A,B\) using the correct basis for your root type. If you choose “Direct constants”, the tool uses your \(A,B\) directly.

Evaluation

Evaluate at \(x^\ast\):

The graph marks \((x_0,y(x_0))\) and \((x^\ast\,y(x^\ast\))) with coordinate labels.

Graph settings (optional)

x min: x max: y min: y max: Plot samples: Engine preference:

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

Ready

Enter \(p,q\) and click “Solve”.

Graph

Solution curve (pan/zoom) — axes/ticks shown

Axes represent \(x\) and \(y\) in your chosen units. The curve updates automatically when you change inputs.

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Theory — Second-Order Homogeneous Linear ODE (Constant Coefficients)

Problem form

A constant-coefficient homogeneous second-order ODE has the form \[ y'' + p\,y' + q\,y = 0, \] where \(p\) and \(q\) are constants. The standard method is to try an exponential solution \(y=e^{rx}\).

Characteristic equation

If \(y=e^{rx}\), then \(y'=re^{rx}\) and \(y''=r^2 e^{rx}\). Substituting into the ODE gives: \[ r^2 e^{rx} + p r e^{rx} + q e^{rx} = 0 \quad\Rightarrow\quad r^2 + p r + q = 0. \] This quadratic is called the characteristic equation.

Root cases and general solution

Let the discriminant be \(\Delta=p^2-4q\).

Distinct real roots (\(\Delta>0\)): \[ r_1=\frac{-p+\sqrt{\Delta}}{2},\quad r_2=\frac{-p-\sqrt{\Delta}}{2}, \qquad y(x)=A e^{r_1 x}+B e^{r_2 x}. \]
Repeated real root (\(\Delta=0\)): \[ r=\frac{-p}{2}, \qquad y(x)=(A+Bx)e^{r x}. \]
Complex conjugate roots (\(\Delta<0\)): \[ r=\alpha\pm i\beta,\quad \alpha=-\frac{p}{2},\quad \beta=\frac{\sqrt{-\Delta}}{2}, \] \[ y(x)=e^{\alpha x}\left(A\cos(\beta x)+B\sin(\beta x)\right). \]

The calculator detects the case automatically and displays the matching solution form.

Using initial conditions

If you are given \(y(x_0)=y_0\) and \(y'(x_0)=v_0\), substitute into the general solution and its derivative to solve for \(A,B\). The result is a 2×2 linear system whose basis depends on the root type.

Example basis functions:

Distinct roots: \(\phi_1=e^{r_1 x},\;\phi_2=e^{r_2 x}\)
Repeated root: \(\phi_1=e^{rx},\;\phi_2=x e^{rx}\)
Complex roots: \(\phi_1=e^{\alpha x}\cos(\beta x),\;\phi_2=e^{\alpha x}\sin(\beta x)\)

Then \(y=A\phi_1+B\phi_2\) and \(y'=A\phi_1'+B\phi_2'\).

Worked example

For \[ y'' + 2y' + 5y = 0, \] the characteristic equation is \(r^2+2r+5=0\). \[ \Delta = 2^2 - 4\cdot 5 = 4-20=-16<0 \Rightarrow r=-1\pm 2i. \] Therefore \[ y(x)=e^{-x}\left(A\cos(2x)+B\sin(2x)\right). \]

University extension: Cauchy–Euler preview

A related class is the Cauchy–Euler equation: \[ x^2 y'' + a x y' + b y = 0, \] which is solved by trying \(y=x^m\) and solving a characteristic equation in \(m\).

Frequently Asked Questions

What is the characteristic equation for y'' + p y' + q y = 0?

Assuming a trial solution y = e^(r x) converts the ODE into the quadratic r^2 + p r + q = 0. Solving this characteristic equation determines the root type and the solution form.

How does the solver handle repeated or complex roots?

If the discriminant p^2 - 4q equals 0, the root is repeated and the solution becomes (A + Bx)e^(r x). If the discriminant is negative, the roots are alpha ± i beta and the solution becomes e^(alpha x)(A cos(beta x) + B sin(beta x)).

How are A and B found from initial conditions?

The solver substitutes y(x0)=y0 and y'(x0)=v0 into the solution basis for the detected root case. This produces a 2x2 linear system that is solved for A and B.

What do underdamped, critically damped, and overdamped mean in this tool?

These presets correspond to the characteristic-root cases: underdamped means complex roots, critically damped means a repeated real root, and overdamped means distinct real roots. The preset buttons provide quick coefficient examples that match each case.

Can the calculator evaluate the solution at a specific x value?

Yes, entering x* computes y(x*) from the solved form and shows the point on the graph alongside the initial-condition point when x0 is provided.