Answer: (Ax + B)e2x.
- A Axe2x
- B (Ax + B)e2x
- C Ae2x
- D (Ax² + Bx)e2x
Correct answer: B. (Ax + B)e2x
Explanation: Rule: for f(x)=pₙ(x)eᵃˣ where pₙ is degree-n polynomial, trial PI=(Ax+B)e2x (same degree polynomial times e2x), provided 2 is not a root of the characteristic equation. Substitute into DE to find A and B.
The general solution y=Ax² represents an entire FAMILY of curves, one for each value of the arbitrary constant A; a particular solution (fixed by an initial condition) selects exactly one curve from this family.
Concept context
Equations involving derivatives and their solutions