Superposition and Fundamental Solutions

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The idea

A linear differential operator respects addition and scalar multiplication. If $L[y_1]=L[y_2]=0$ , then $c_1y_1+c_2y_2$ also solves the homogeneous equation. For a second-order equation, two independent solutions form a fundamental pair.

Visual guide

VISUAL GUIDECombine independent modes to set the initial state

cos t
sin t
2 cos t − 3 sin t

Cos t and sin t solve y″ + y = 0. Their combination 2 cos t − 3 sin t also solves it and has y(0) = 2, y′(0) = −3. The combination changes amplitude and phase, not the equation.

Method and assumptions

For $y''+p(t)y'+q(t)y=0$ with continuous coefficients, test independence with the Wronskian $W=y_1y_2'-y_1' y_2$ . If it is nonzero at one point, the pair spans every solution on that interval. For $L[y]=g$ , write $y=y_h+y_p$ using any one particular solution.

Worked example: oscillation basis

For $y''+y=0$ , choose $\cos t,\sin t$ . Their Wronskian is $1$ . Conditions $y(0)=2,y'(0)=-3$ select $y=2\cos t-3\sin t$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For a second-order homogeneous linear equation, two solutions form a fundamental pair when they are…

Identical

Both constant

Linearly independent

Hint 1 · Find a starting point

The general solution needs two independent degrees of freedom.

Hint 2 · Take the next step

One solution must not be a constant multiple of the other.

Show the reasoning

Answer: Linearly independent

Independent solutions span the two-parameter solution family on a regular interval.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: dependent candidates

The functions $e^t$ and $2e^t$ both solve $y''-y=0$ , but their Wronskian is zero. Their combinations cannot produce $e^{-t}$ . The independent pair $e^t,e^{-t}$ is required to describe all initial states.

Interpreting the result

Superposition applies to homogeneous linear equations. Adding two solutions with the same nonzero forcing doubles that forcing. Nonlinear equations generally do not allow superposition at all.

Practice

Find the Wronskian of $1,t$ .
Solve $y''=0$ with $y(0)=1,y'(0)=2$ .
If $L[y_p]=g$ , what is $L[3y_p]$ ?

Show worked solutions

$W=1$ , so they are independent.
$y=1+2t$ .
$3g$ by linearity, not $g$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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