Mathwords logoMathwords

Superposition Principle — Definition, Formula & Examples

The superposition principle is the rule that if you have two solutions to a linear homogeneous differential equation, then any constant multiple of each, added together, is also a solution. It lets you build new solutions from ones you already know.

Let LL be a linear differential operator, so that L[c1y1+c2y2]=c1L[y1]+c2L[y2]L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2] for all constants c1,c2c_1, c_2 and all sufficiently differentiable functions y1,y2y_1, y_2. If L[y1]=0L[y_1] = 0 and L[y2]=0L[y_2] = 0, then L[c1y1+c2y2]=0L[c_1 y_1 + c_2 y_2] = 0. More generally, if L[y1]=g1L[y_1] = g_1 and L[y2]=g2L[y_2] = g_2, then L[c1y1+c2y2]=c1g1+c2g2L[c_1 y_1 + c_2 y_2] = c_1 g_1 + c_2 g_2.

Key Formula

L[y1]=0,  L[y2]=0  ⟹  L[c1y1+c2y2]=0,where L[y]=y′′+p(x) y′+q(x) yL[y_1] = 0,\; L[y_2] = 0 \;\Longrightarrow\; L[c_1 y_1 + c_2 y_2] = 0, \quad \text{where } L[y] = y'' + p(x)\,y' + q(x)\,y
Where:
  • LL = A linear differential operator. The second-order operator shown is one common case.
  • y1,y2y_1, y_2 = Known solutions of the homogeneous equation L[y] = 0
  • c1,c2c_1, c_2 = Arbitrary constants
  • p(x),q(x)p(x), q(x) = Coefficient functions, which do not depend on y

Worked Example

Problem: The functions y1=exy_1 = e^x and y2=e−xy_2 = e^{-x} both solve y′′−y=0y'' - y = 0. Use the superposition principle to find another solution. Then verify it directly.
Step 1: Check the given solutions. Each exponential is its own second derivative up to sign, so y1′′=exy_1'' = e^x and y2′′=e−xy_2'' = e^{-x}. Both satisfy the equation.
y1′′−y1=ex−ex=0,y2′′−y2=e−x−e−x=0y_1'' - y_1 = e^x - e^x = 0, \qquad y_2'' - y_2 = e^{-x} - e^{-x} = 0
Step 2: The equation is linear and homogeneous, so superposition applies. Choose c1=2c_1 = 2 and c2=3c_2 = 3, then differentiate the combination twice.
y=2ex+3e−x,y′=2ex−3e−x,y′′=2ex+3e−xy = 2e^x + 3e^{-x}, \quad y' = 2e^x - 3e^{-x}, \quad y'' = 2e^x + 3e^{-x}
Step 3: Substitute into the left side of the equation.
y′′−y=(2ex+3e−x)−(2ex+3e−x)=0y'' - y = (2e^x + 3e^{-x}) - (2e^x + 3e^{-x}) = 0
Answer: y=2ex+3e−xy = 2e^x + 3e^{-x} is a solution of y′′−y=0y'' - y = 0. By the same reasoning, so is y=c1ex+c2e−xy = c_1 e^x + c_2 e^{-x} for every choice of c1c_1 and c2c_2.

Why It Matters

Superposition is why the general solution of a second-order linear homogeneous equation has the form c1y1+c2y2c_1 y_1 + c_2 y_2, where y1y_1 and y2y_2 are linearly independent solutions. You then fit c1c_1 and c2c_2 to initial or boundary conditions. Engineers use the same idea in circuit analysis, vibration problems, and Fourier series solutions of the heat and wave equations.

Common Mistakes

Mistake: Applying superposition to nonhomogeneous or nonlinear equations. For example, adding two solutions of y′′+y=1y'' + y = 1 and expecting the sum to solve y′′+y=1y'' + y = 1.
Correction: If L[y1]=1L[y_1] = 1 and L[y2]=1L[y_2] = 1, then L[y1+y2]=2L[y_1 + y_2] = 2, not 1. For a nonhomogeneous equation, the correct method adds one particular solution to the general homogeneous solution. Superposition fails completely for nonlinear equations such as y′=y2y' = y^2.

Related Terms