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Delta Functions and Forcing

Impulse forcing like a hammer blow, modeled by the Dirac delta, and solving initial value problems driven by an input concentrated at one instant.

The ODE Project, Delta Functions and Forcing

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6 exercises

Exercise 1

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

2y′′+y′+2y=δ(t)2 y'' + y' + 2y = \delta(t), y(0)=0y(0) = 0, y′(0)=0y'(0) = 0

2.5

Exercise 2

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

y′′−y′−2y=δ(t−5)y'' - y' - 2y = \delta(t - 5), y(0)=0y(0) = 0, y′(0)=0y'(0) = 0

2.5

Exercise 3

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

d2xdx2−6dxdt+25x=δ(t)\dfrac{d^2x}{dx^2} - 6 \dfrac{dx}{dt} + 25 x = \delta(t), x(0)=1x(0) = 1, x′(0)=−2x'(0) = -2

2.5

Exercise 4

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

y′′+16y=δ(t)y'' + 16y = \delta(t), y(0)=1y(0) = 1, y′(0)=0y'(0) = 0

2.5

Exercise 5

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

y′′+16y=δ(t−4)y'' + 16y = \delta(t - 4), y(0)=1y(0) = 1, y′(0)=0y'(0) = 0

2.5

Exercise 6

Solve the initial problems in Solving Initial Value Problems using the Laplace transform, δ(t)\delta(t) is the unit impulse function.

y′′+2y′+y=δ(t+4)y'' + 2y' + y = \delta(t + 4), y(0)=−1y(0) = -1, y′(0)=3y'(0) = 3

2.5

Adapted from The ODE Project © 2019–2025 Thomas W. Judson. Licensed under CC BY 4.0. Source: The ODE Project. Imported at revision 7935458.