# Initial-value problems and equilibrium

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

Course: [Differential equations and numerical methods](https://studybrix.com/en/courses/differential-equations/index.md)
Level: university
Prerequisites: Differentiation, exponentials and integration.

## 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.

## Practice questions

### 1. Solve y′=3y with y(0)=2.

Sample answer / marking guidance: y(t)=2e^(3t).

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

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

Answer: 4

Explanation: Set y′=0 so 2y=8.

### 3. Solve y′+2y=8 with y(0)=1.

Sample answer / marking guidance: y(t)=4−3e^(−2t).

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

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

A. They become instantly zero
B. They decay exponentially
C. They grow exponentially
D. They oscillate without decay

Answer: They decay exponentially

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

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

Answer: 7

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

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

Sample answer / marking guidance: Infinitely many functions pass through the initial point. The function must also satisfy the differential equation throughout its interval.

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

### 7. Which form is the homogeneous part of y′+ay=b for a>0?

A. A fixed value zero for all initial data
B. Ce^(−at)
C. Ce^(at)
D. bt only

Answer: Ce^(−at)

Explanation: The homogeneous equation is y′+ay=0.

### 8. Explain why dividing by y when solving y′=y can hide a solution.

Sample answer / marking guidance: Division assumes y≠0. The constant solution y=0 must be checked separately, though it is recovered by allowing C=0 in the final exponential family.

Explanation: Credit the excluded case and direct verification.

## Further reading

- https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/
- https://ocw.mit.edu/courses/18-06-linear-algebra-spring-2010/

Original AI-assisted educational draft; not independently teacher-reviewed. References support further study, not copied exercises, affiliation or curriculum accreditation.