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MathidiaMathematics Academy
Differential equations II
Lesson 29 of 75
Challenge

Laplace transforms

Objective. Solve initial-value and discontinuous forcing problems.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Laplace transforms convert a linear ODE in t into an algebraic equation in s via L{f(t)} = ∫0^∞ f(t)e^(−st) dt, turning derivatives into multiplication by s. This makes initial-value problems straightforward: L{y′} = sY(s) − y(0), so an ODE with given initial data becomes an algebraic equation for Y(s), which is then inverted using a transform table. The method handles discontinuous or impulsive forcing naturally, using the unit step function u(t−a) with L{u(t−a)f(t−a)} = e^(−as)F(s) and the delta function for impulses.

Worked example

Use a Laplace transform to solve y′ + 2y = 0 with y(0) = 5.

2 … 0
Coordinate grid with the line 2 … 0
  1. 1Transform: sY(s) − y(0) + 2Y(s) = 0
  2. 2Substitute y(0) = 5 to get (s+2)Y(s) = 5
  3. 3Solve for Y(s) = 5/(s+2)
  4. 4Invert using L⁻¹{1/(s+a)} = e^(−at)

Answer: y(t) = 5e^(−2t)

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Common mistake. Students often forget to include the initial conditions when transforming derivatives, writing L{y′} = sY(s) instead of sY(s) − y(0). Always substitute the given initial values immediately when transforming.

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