First-Order Linear Equations

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The idea

For $y'+p(t)y=q(t)$ , the integrating factor converts the left side to one product derivative. This works because the factor is selected to satisfy $\mu'=p\mu$ . It avoids guessing a solution for each forcing function.

Visual guide

VISUAL GUIDEThe initial deviation decays toward equilibrium

y(0) = 1
y(0) = 5
Equilibrium y = 3

For y′ + 2y = 6, all solutions approach y = 3 in forward time. The curves starting at 1 and 5 have equal and opposite deviations from equilibrium, each multiplied by e⁻²ᵗ.

Method and assumptions

On an interval with continuous $p,q$ , choose $\mu=e^{\int p(t)dt}$ . Then $(\mu y)'=\mu q$ , so $y=\mu^{-1}(\int\mu q\,dt+C)$ . A nonzero constant multiple of $\mu$ gives the same family. Normalize the coefficient of $y'$ first.

Worked example: constant input

$y'+2y=6$ has $\mu=e^{2t}$ . Integration gives $e^{2t}y=3e^{2t}+C$ , so $y=3+Ce^{-2t}$ . With $y(0)=1$ , $y=3-2e^{-2t}$ approaches the steady value $3$ .

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 y′+2y=t, which integrating factor is standard?

e²ᵗ

eᵗ²

Hint 1 · Find a starting point

Use exp(∫p(t)dt) for y′+p(t)y=q(t).

Hint 2 · Take the next step

The coefficient of y is the constant 2.

Show the reasoning

Answer: e²ᵗ

The factor e²ᵗ makes the left side the derivative of e²ᵗy.

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

Worked example: variable coefficients

On $t>0$ , solve $y'+y/t=t$ with $y(1)=2$ . The factor is $t$ , giving $(ty)'=t^2$ . Hence $y=t^2/3+C/t$ and $C=5/3$ . The solution interval is constrained by the coefficient singularity at zero.

Interpreting the result

The complementary part carries initial-condition memory; the forced part reflects input. Their long-time behavior depends on the coefficient sign. A positive constant damping coefficient erases initial differences exponentially.

Practice

Solve $y'+y=0$ , $y(0)=5$ .
Find the integrating factor for $y'+2ty=1$ .
Find the steady value for $y'+4y=8$ .

Show worked solutions

$y=5e^{-t}$ .
$\mu=e^{t^2}$ ; the remaining integral need not have an elementary antiderivative.
The equilibrium is $y=2$ . All solutions approach it for forward time.

Further study

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

MAKE IT YOURS

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