Differential Equations and Solution Families

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The idea

An ordinary differential equation relates an unknown function of one variable to its derivatives. A solution is a differentiable function satisfying that relation on an interval. The order is the highest derivative present. A second-order equation usually needs two independent initial conditions to select a particular solution.

Visual guide

VISUAL GUIDEOne equation, a family of solutions

C = 3
C = 1
C = 0
C = -2

Every curve Ce⁻²ᵗ has slope −2y at each point, so all solve the same equation y′ = −2y. Changing C selects a different initial value; C = 0 gives the equilibrium.

Method and assumptions

Identify the independent variable, unknown and units. An equation is linear when the unknown and its derivatives enter to the first power without products between them; coefficients may depend on the independent variable. Verify candidates by differentiating and substituting, then state the interval on which every expression exists.

Worked example: a family of cooling curves

For $y'=-2y$ , the functions $y=Ce^{-2t}$ satisfy $y'=-2Ce^{-2t}$ . The constant is arbitrary until an initial value is supplied. The zero function belongs to the family with $C=0$ .

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.

Which function solves y′=2y for every real t?

y=2t+3

y=3e²ᵗ

y=3eᵗ

Hint 1 · Find a starting point

Differentiate each candidate and compare with twice its value.

Hint 2 · Take the next step

The derivative of e²ᵗ introduces a factor 2.

Show the reasoning

Answer: y=3e²ᵗ

For y=3e²ᵗ, y′=6e²ᵗ=2y. A solution must satisfy the equation, not just an initial value.

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

Worked example: finite-time growth

For $y'=y^2$ , $y=1/(1-t)$ works because its derivative is $1/(1-t)^2$ . With $y(0)=1$ , its maximal interval containing zero is $(-\infty,1)$ . The expression also exists beyond $1$ , but cannot pass through its singularity as one continuous solution.

Interpreting the result

An algebraic expression without a valid interval is an incomplete answer. Nonlinear equations can have finite-time blow-up even when their formulas look elementary.

Practice

Classify $y''+t y=\sin t$ .
Does $y=3e^t$ solve $y'=2y$ ?
Give an equilibrium solution of $y'=y^2$ .

Show worked solutions

It is a second-order linear nonhomogeneous equation.
No: the derivative is $3e^t$ , whereas $2y=6e^t$ .
$y=0$ has zero derivative and satisfies the equation.

Further study

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

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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