Differential equations

The app solves first-order ordinary differential equations: separable, linear, exact, homogeneous (y′ = f(y/x)), Bernoulli equations and equations y′ = f(ax + by + c). It recognizes the type itself and shows the steps as taught at school – separating the variables, variation of the constant or an integrating factor, a substitution, the potential of an exact equation. The integrals in the steps have steps of their own.

Higher orders: linear equations with constant coefficients of any order by the characteristic equation (multiple and complex roots too), a particular solution by the special right side (resonance and a sum of right sides too) or by variation of constants, an initial value problem at zero by the Laplace transform too (the "Method" chip). Further Euler's equation by x = eᵗ, y⁽ⁿ⁾ = f(x) by repeated integration and the reduction of order – when y is missing (p = y′) or x is (y′ = p(y)).

The equation y′ = y·(a(x)·ln y + b(x)) becomes linear by the substitution z = ln y. A linear second-order equation with variable coefficients is solved when one solution can be guessed (xᵏ, eᶜˣ, …): the second one by reducing the order (d'Alembert), the right side by variation of constants. The order is reduced by more than one too (p = y⁽ᵐ⁾ with a second-order equation for p). With parameters the cases are told apart – a damped oscillator by the sign of δ² − ω² (overdamped, critical, underdamped), a resonance (y″ + ω²y = cos 2t for ω = 2) and the sign of a constant the integral depends on (y″ = 2yy′). The Laplace transform also solves an initial value problem at a point other than 0 (by the shift τ = t − t₀) and with parameters.

The result is the general solution with a constant C, for higher orders with constants C₁, C₂, … (explicitly, or implicitly where it can't be explicit); with conditions, the solution that satisfies them – initial and boundary ones, including "no solution" and "infinitely many solutions". Constant solutions lost by dividing are found too. Every solution is checked by substituting it into the equation. For the first order the graph shows the direction field and solutions for different C, for higher orders solutions for different constants; for an autonomous second-order equation (no x) a switch shows the phase plane (y, y′) too.

How to enter it

y' = x*yThe derivative with a prime.
dy/dx = x y^2Leibniz's notation – it names the variable too (dm/dt, dT/dt).
\frac{dy}{dx} = \frac{y}{x} + xNotation from a photo.
(2xy + 6x)dx + (x^2 + 4y^3)dy = 0The form M dx + N dy = 0 (an exact equation).
y' = y^2 + 1, y(0) = 1An initial condition after a comma or on the next line.
dm/dt = m + 2, m(0) = 0The function and the variable can be any letters.
y'' - 3y' + 2y = 3e^(2x)A higher-order equation – the derivatives with more primes.
y^{(4)} + 2y'' + y = 0Higher derivatives as y^{(4)} (from a photo too).
y'' + y = 0, y(0) = 1, y'(0) = 2Initial conditions y(0), y′(0) – or boundary ones at different points.
\ddot{x} + \omega^2 x = 0Newton's dots (derivatives by time t) and Greek parameters.

Examples

A separable equation – the general solution and the direction field.
Input: y' = x*y
y' = x*y
y' = x*yy' = x*y
The initial condition determines the constant C.
Input: y' = y^2 + 1, y(0) = 1
y' = y^2 + 1, y(0) = 1
y' = y^2 + 1, y(0) = 1y' = y^2 + 1, y(0) = 1y' = y^2 + 1, y(0) = 1
A linear equation by variation of the constant.
Input: y' + 2y = e^x Variation of the constant
y' + 2y = e^x
y' + 2y = e^xy' + 2y = e^x
The same equation by an integrating factor.
Input: y' + 2y = e^x Integrating factor
y' + 2y = e^x
y' + 2y = e^xy' + 2y = e^x
A homogeneous equation – the substitution u = y/x.
Input: y' = (x - y)/(x + y)
y' = (x - y)/(x + y)
y' = (x - y)/(x + y)y' = (x - y)/(x + y)y' = (x - y)/(x + y)
A Bernoulli equation – the substitution z = y^(1−n) and the constant solution y = 0.
Input: xy' - y = -xy^2
xy' - y = -xy^2
xy' - y = -xy^2xy' - y = -xy^2xy' - y = -xy^2
An exact equation – the potential F(x, y) = C.
Input: (2xy + 6x)dx + (x^2 + 4y^3)dy = 0
(2xy + 6x)dx + (x^2 + 4y^3)dy = 0
(2xy + 6x)dx + (x^2 + 4y^3)dy = 0(2xy + 6x)dx + (x^2 + 4y^3)dy = 0(2xy + 6x)dx + (x^2 + 4y^3)dy = 0
The substitution z = x − y; Symja would lose a factor 2 here, the app checks the result.
Input: y' = cos(x - y), y(0) = pi/2
y' = cos(x - y), y(0) = pi/2
y' = cos(x - y), y(0) = pi/2y' = cos(x - y), y(0) = pi/2
The characteristic equation with complex roots −2 ± 3i.
Input: y'' + 4y' + 13y = 0
y'' + 4y' + 13y = 0
y'' + 4y' + 13y = 0y'' + 4y' + 13y = 0
A special right side with resonance – 2 is a root of the characteristic equation.
Input: y'' - 3y' + 2y = 3e^(2x) Special right side
y'' - 3y' + 2y = 3e^(2x)
y'' - 3y' + 2y = 3e^(2x)y'' - 3y' + 2y = 3e^(2x)
The right side tan x isn't special – variation of constants and the Wronskian.
Input: y'' + y = tan x
y'' + y = tan x
y'' + y = tan xy'' + y = tan xy'' + y = tan x
An initial value problem – a system for the constants C₁, C₂.
Input: y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0
y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0
y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0
The same problem by the Laplace transform.
Input: y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0 Laplace transform
y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0
y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0y'' - 4y' + 3y = e^(2x), y(0) = 1, y'(0) = 0
A boundary value problem with infinitely many solutions.
Input: y'' + y = 0, y(0) = 0, y(pi) = 0
y'' + y = 0, y(0) = 0, y(pi) = 0
Euler's equation – the substitution x = eᵗ.
Input: x^2y'' - 2xy' + 2y = x^2 + 2
x^2y'' - 2xy' + 2y = x^2 + 2
x^2y'' - 2xy' + 2y = x^2 + 2x^2y'' - 2xy' + 2y = x^2 + 2x^2y'' - 2xy' + 2y = x^2 + 2
The reduction y′ = p(y), the constants from the initial conditions step by step.
Input: y'' = e^(2y), y(0) = 0, y'(0) = 1
y'' = e^(2y), y(0) = 0, y'(0) = 1
y'' = e^(2y), y(0) = 0, y'(0) = 1y'' = e^(2y), y(0) = 0, y'(0) = 1y'' = e^(2y), y(0) = 0, y'(0) = 1
A damped oscillator – three cases by δ and ω.
Input: y'' + 2δy' + ω^2y = 0
y'' + 2δy' + ω^2y = 0
Resonance – for ω = 2 the solution has a different form.
Input: y'' + ω^2y = cos(2t)
y'' + ω^2y = cos(2t)
y'' + ω^2y = cos(2t)y'' + ω^2y = cos(2t)
A guessed solution y₁ = x and reduction of order (d'Alembert).
Input: (1-x^2)y'' - 2xy' + 2y = 0
(1-x^2)y'' - 2xy' + 2y = 0
(1-x^2)y'' - 2xy' + 2y = 0(1-x^2)y'' - 2xy' + 2y = 0(1-x^2)y'' - 2xy' + 2y = 0

Known limitations