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Mathematics · University

Initial-value problems and equilibrium

Solving a differential equation produces constants; initial conditions select a particular solution.

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Prerequisites

Differentiation, exponentials and integration.

Study notes

General versus particular

Solving a differential equation produces constants; initial conditions select a particular solution.

Equilibrium

A constant solution is found by setting the derivative to zero.

Existence and uniqueness

Conditions on the right-hand side matter; continuity alone does not always imply uniqueness.

Verification

Differentiate a proposed solution and check both the equation and the initial condition.

Worked examples

Solve y′=3y with y(0)=2.

Credit a general exponential solution, applying y(0), and correct substitution.

y(t)=2e^(3t).

For y′+2y=8, find the equilibrium solution value.

Set y′=0 so 2y=8.

4

Check your understanding

  1. Solve y′+2y=8 with y(0)=1.

  2. For y′=−5y, what happens to small deviations from y=0 as t increases?

    1. They become instantly zero
    2. They decay exponentially
    3. They grow exponentially
    4. They oscillate without decay

  3. For y′=2t and y(0)=3, calculate y(2).

  4. Why does checking only the initial condition fail to verify a proposed solution?

Answers and marking guidance

Answers and marking notes
  1. y(t)=4−3e^(−2t).

    Credit the equilibrium plus a decaying homogeneous solution and the initial condition.

  2. They decay exponentially

    The solution is Ce^(−5t), so the equilibrium is asymptotically stable.

  3. 7

    Integration gives y=t²+3, so y(2)=7.

  4. Infinitely many functions pass through the initial point. The function must also satisfy the differential equation throughout its interval.

    Credit the distinction between a pointwise constraint and a differential relation.

Further reading and review references

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