Linear PDE Calculator
Solve linear PDEs with constant coefficients, f(D, D′)z = F(x, y): homogeneous operators by the auxiliary equation, non-homogeneous ones (with D, D′ or a constant term) by linear factors, and the particular integral for exponential, sine, cosine and polynomial right sides, including the cases where the usual denominator is 0.
About the Linear PDE Calculator
Putting D = mD′ turns the operator into Am² + Bm + C. Distinct roots m₁, m₂ give the complementary function φ₁(y + m₁x) + φ₂(y + m₂x); a repeated root m gives φ₁(y + mx) + xφ₂(y + mx).
A non-homogeneous operator that factors as (α₁D + β₁D′ + γ₁)(α₂D + β₂D′ + γ₂) has CF e^(−γ₁x/α₁)φ₁(α₁y − β₁x) + e^(−γ₂x/α₂)φ₂(α₂y − β₂x). For F = e^(px + qy), the particular integral is e^(px + qy)/f(p, q), or x·e^(px + qy)/f_D(p, q) when f(p, q) = 0. For sin or cos(px + qy), replace D² by −p², DD′ by −pq and D′² by −q². For a polynomial F, expand 1/f(D, D′) as a series in D′/D, which stops after a few terms.
The formula
D = mD′: Am² + Bm + C = 0; z = φ₁(y + m₁x) + φ₂(y + m₂x) + PI
Worked example
(D² − 5DD′ + 6D′²)z = e^(x + y)
- m² − 5m + 6 = 0: m = 2, 3; f(1, 1) = 1 − 5 + 6 = 2
- z = φ₁(y + 2x) + φ₂(y + 3x) + e^(x + y)/2
Things to keep in mind
- Operators up to second order. A non-homogeneous operator whose second-order part doesn't split into rational linear factors gets the CF in the form Σ A e^(hx + ky) with f(h, k) = 0. Lagrange's linear equation and Charpit's method have their own pages.
Questions people ask
Why arbitrary functions instead of constants?
A PDE's general solution has arbitrary functions: any φ(y + mx) satisfies (D − mD′)φ = 0.
Does the method work when the auxiliary roots are complex?
Yes. The complementary function is still φ₁(y + m₁x) + φ₂(y + m₂x) with complex m; real solutions come from combining the two.