Variation of Parameters

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The idea

Variation of parameters replaces the constants in a homogeneous solution by functions. Unlike undetermined coefficients, it handles general forcing and variable coefficients, provided a fundamental homogeneous pair is known.

Visual guide

VISUAL GUIDEA particular solution leaves homogeneous freedom

yₚ
yₚ + 1
yₚ + t

For y″ = 1/t on t > 0, yₚ = t ln t − t is one particular solution. Adding 1 or adding t changes the initial data without changing the second derivative, because both additions solve y″ = 0.

Method and assumptions

Normalize to $y''+p y'+q y=g$ . With independent $y_1,y_2$ and $W=y_1y_2'-y_1' y_2$ , impose $u_1' y_1+u_2' y_2=0$ . The remaining equation gives $u_1'=-y_2g/W$ and $u_2'=y_1g/W$ . Integrate and set $y_p=u_1y_1+u_2y_2$ .

Worked example: an elementary check

For $y''+y=1$ , use $y_1=\cos t,y_2=\sin t,W=1$ . Then $u_1=\cos t,u_2=\sin t$ are convenient antiderivatives. Therefore $y_p=\cos^2t+\sin^2t=1$ .

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.

Why must the fundamental pair’s Wronskian be nonzero in variation of parameters?

It forces the particular solution to vanish.

It means the forcing is constant.

It makes the equations for the varying coefficients solvable.

Hint 1 · Find a starting point

The coefficient derivatives satisfy a 2×2 linear system.

Hint 2 · Take the next step

The Wronskian is that system’s determinant.

Show the reasoning

Answer: It makes the equations for the varying coefficients solvable.

A nonzero determinant permits solving for the two coefficient derivatives; it encodes independence.

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

Worked example: non-polynomial forcing

For $y''=1/t$ on $t>0$ , choose $y_1=1,y_2=t,W=1$ . Then $u_1'=-1,u_2'=1/t$ , giving $y_p=-t+t\ln t$ . Differentiating twice returns $1/t$ .

Interpreting the result

Changing integration constants only adds homogeneous terms, so one convenient set is enough for a particular solution. Definite integrals from an initial time are useful for enforcing zero initial response and for numerical evaluation.

Practice

Why must $W$ be nonzero?
What must be done to $2y''+2y=f(t)$ first?
Find a particular solution of $y''=e^t$ .

Show worked solutions

The coefficient system for $u_1',u_2'$ must be invertible.
Divide by $2$ , so the normalized forcing is $g=f/2$ .
$y_p=e^t$ ; differentiating twice verifies it. The general solution adds $c_1+c_2t$ .

Further study

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

MAKE IT YOURS

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