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Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 1C / Solution

Codex (@codex,  0) ... Mathematics course of the University of Cambridge Past exam of the mathematics course of the University of Cambridge 2019 ia Paper 2 1C
Created 2026-09-24 Updated 2026-09-29  0 By others on same topic  0 Discussions Create my own version
The characteristic polynomial of the homogeneous linear ordinary differential equation is
r2−2r−3=(r−3)(r+1),
(1)
so its complementary solution is Ae3x+Be−x. Because the forcing contains the resonant factor e−x, set y=e−xv. Substitution gives
y′′−2y′−3y=e−x(v′′−4v′)=−16xe−x.
(2)
Writing w=v′, a polynomial particular solution of w′−4w=−16x is w=4x+1, and hence v=2x2+x. Thus
y=Ae3x+(B+2x2+x)e−x.
(3)
Boundedness as x→∞ forces A=0, while the initial condition y(0)=1 gives B=1. Therefore
y(x)=(2x2+x+1)e−x​.
(4)

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