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Homogeneous solution forms

Write the real general solution from characteristic roots.

Differential Equations · Second order
$$r_1\ne r_2:\ y=C_1e^{r_1x}+C_2e^{r_2x};\quad r:\ y=(C_1+C_2x)e^{rx};\quad \alpha\pm i\beta:\ y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$$

Homogeneous solution forms is one of 2 second order formulas in the differential equations section of this library, and it is used at university level.

Why homogeneous solution forms works

A second-order equation needs two independent building blocks, and the characteristic roots supply them. Two different roots give two different exponentials. A repeated root gives only one, so multiplying by x manufactures a second that is genuinely independent. A complex pair gives exponentials with imaginary exponents, and Euler's formula recombines them into a real sine and cosine.

What each symbol means

$r_1,r_2$ are roots and $C_1,C_2$ arbitrary constants.

Homogeneous solution forms: when it holds

This applies to second-order homogeneous linear equations with constant coefficients.

When it stops applying

These three shapes cover second-order homogeneous equations with constant coefficients and nothing else. A forcing term on the right needs a particular solution added on, and coefficients that depend on x change the whole method, since then the roots are no longer constants.

Homogeneous solution forms: a worked example

Roots $\pm i$ give $y=C_1\cos x+C_2\sin x$.

The mistake to avoid

What people do: Students handle a repeated root by writing y = (C₁ + C₂)e^(rx).

Why it goes wrong: Adding two constants together just makes one constant, so that expression contains a single unknown and cannot meet two initial conditions.

Do this instead: Attach the extra x to the second term: y = (C₁ + C₂x)e^(rx). For y'' − 4y' + 4y = 0 with the double root 2, substituting y = xe^(2x) does satisfy the equation, which is the check worth doing once.

Homogeneous solution forms: step by step

  1. Name the unknown, and the unit the answer has to come out in.
  2. Match the symbols to your values. $r_1,r_2$ are roots and $C_1,C_2$ arbitrary constants.
  3. Check the conditions before substituting. This applies to second-order homogeneous linear equations with constant coefficients.
  4. Substitute, keep exact values to the last line, then test the sign, size, and unit against a rough estimate — the check that catches most differential equations slips.

Where this formula fits

Subject
Differential Equations formulas — 8 entries in this library
Topic
Second order
Level
University

Formulas are easiest to keep when they sit inside a method rather than on a list. Use the links below to see where homogeneous solution forms comes from, to check a calculation against a tool, and to practise it until you can recall it without looking.

Questions about homogeneous solution forms

Why does a repeated root need an extra x?

Because the two solutions must be independent, and one exponential used twice is not. Multiplying by x tilts the second copy just enough that no constant multiple of the first can reproduce it.

Where did the i go in the complex case?

It got absorbed by Euler's formula. Combining the two conjugate exponentials makes the imaginary parts cancel, leaving e^(αx) times a cosine and a sine with real coefficients.

Do I use both roots when they are a conjugate pair?

You use the pair once, not twice. The single pair α ± iβ already produces both building blocks, cos βx and sin βx, so writing four terms would double count.

How do I pin down C₁ and C₂?

With two initial conditions, usually the value of y and of y' at one point. Substituting them gives two equations in the two constants, which you solve as an ordinary system.

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