Superposition · Differential Equations

Intuition

For a homogeneous linear equation, solutions can be added and multiplied by constants and they remain solutions. This is superposition, and it is why linear equations are tractable: find two independent solutions and every combination of them is available. It fails the moment the equation is forced or nonlinear, and the ways it fails are as instructive as the principle.

Two ripples on a pond pass through each other and add, each carrying on afterwards as if the other had not been there. Small water waves obey a linear equation, and superposition is what the eye sees.

For $y''-y=0$ , $y_{1}=e^{x}$ and $y_{2}=e^{-x}$ are solutions, and so is any combination: here their average $\tfrac12(e^{x}+e^{-x})=\cosh x$ . Every solution is $C_{1}e^{x}+C_{2}e^{-x}$ .

Combinations of solutions are solutions

Write $L[y]=y''+p(x)y'+q(x)y$ . Then for any functions and constants: is linear in the sense of the linear algebra course. So the solutions of form a vector space of functions, and every solution of the forced equation is one particular solution plus a solution of .

L[y_{1}]=L[y_{2}]=0\ \Longrightarrow\ L[c_{1}y_{1}+c_{2}y_{2}]=0

Where it holds, and where it fails

$\cos x$ and $\sin x$ solve $y''+y=0$ , so does too.

The superposition principle

L[y_{1}]=0,\ L[y_{2}]=0\ \Longrightarrow\ L[c_{1}y_{1}+c_{2}y_{2}]=0

Differentiation is linear: the derivative of a combination is the same combination of the derivatives, once and then again. Multiplying by p or by q distributes over the combination. So L of the combination is the combination of L of each function, and each of those is zero.

Proof steps

\begin{aligned}(c_{1}y_{1}+c_{2}y_{2})'&=c_{1}y_{1}'+c_{2}y_{2}'\\(c_{1}y_{1}+c_{2}y_{2})''&=c_{1}y_{1}''+c_{2}y_{2}''\end{aligned}

Differentiation is linear, once and twice.

\begin{aligned}L[c_{1}y_{1}+c_{2}y_{2}]&=c_{1}\bigl(y_{1}''+py_{1}'+qy_{1}\bigr)\\&\quad+c_{2}\bigl(y_{2}''+py_{2}'+qy_{2}\bigr)\end{aligned}

Substitute and collect the terms belonging to each function.

L[c_{1}y_{1}+c_{2}y_{2}]=c_{1}L[y_{1}]+c_{2}L[y_{2}]

Each bracket is L applied to one of the two functions.

c_{1}\cdot0+c_{2}\cdot0=0

Both functions solve the homogeneous equation, so the combination does too.

Applications

Practice

Add and Scale

Solutions of a homogeneous linear equation can be added and multiplied by constants, and remain solutions.

L[c_{1}y_{1}+c_{2}y_{2}]=c_{1}L[y_{1}]+c_{2}L[y_{2}]

Try it

Given that $\cos2x$ and $\sin2x$ solve $y''+4y=0$ , which of these also solves it?