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Second-Order Linear Homogeneous ODE

Second-order homogeneous ODE solver. Enter a, b to find characteristic roots and write the general solution for three root cases.

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y″ + a y′ + b y = 0, r² + a r + b = 0
real number
real number

📖 Tutorial｜Second-Order Linear Homogeneous ODE

1. Definition
The standard form is y″ + a y′ + b y = 0. Its general solution is determined by the roots of the characteristic equation r² + ar + b = 0.
2. Symbols
SymbolMeaning
a, bconstant coefficients
rcharacteristic root
Δdiscriminant a²−4b
C₁,C₂arbitrary constants
3. How this tool works
  • Set y=e^(rx); the characteristic equation is r²+ar+b=0;
  • Δ>0: two distinct real roots, y=C₁e^(r₁x)+C₂e^(r₂x);
  • Δ=0: repeated root r, y=(C₁+C₂x)e^(rx);
  • Δ<0: conjugate roots α±βi, y=e^(αx)(C₁cos βx+C₂sin βx).
4. Steps
  1. Enter a (coefficient of y′) and b (coefficient of y);
  2. Click Calculate to see the discriminant and roots;
  3. Read the general solution for your case.
5. Example
Example: y″+3y′+2y=0. a=3,b=2, Δ=9−8=1>0, r₁=−1, r₂=−2; general solution y=C₁e^(−x)+C₂e^(−2x).
6. Pitfalls
Put the equation in standard form y″+ay′+by=0 (coefficient of y″ = 1);
For a repeated root the second solution is xe^(rx), not e^(rx);
For complex roots use both the real part α and imaginary part β.

❓ FAQ｜Second-Order Linear Homogeneous ODE

Where does the characteristic equation come from?
Assume y=e^(rx), substitute, and cancel e^(rx) to get r²+ar+b=0.
How to tell the three cases?
By the discriminant Δ=a²−4b: positive two real roots, zero repeated root, negative conjugate complex roots.
Why multiply by x for a repeated root?
e^(rx) alone gives one independent solution; the second is xe^(rx) by reduction of order.
Why sin and cos for complex roots?
The real and imaginary parts of e^((α+βi)x) are e^(αx)cos βx and e^(αx)sin βx.
What if the coefficient of y″ is not 1?
Divide through by it first to obtain the standard form.