Solving Initial Value Problems with Laplace Transforms
Transform the equation, solve the resulting algebra for the transform of the unknown, then invert. Initial conditions enter at the start rather than at the end, which is what makes the method suit discontinuous forcing.
What a learner can do afterwards
- Transform an initial value problem and solve for the transform of the solution
- Invert with partial fractions and a table
- Handle a forcing term that switches on at a fixed time
1 · Read
The Laplace transform trades calculus for algebra. Each derivative becomes multiplication by s, with the starting values peeled off as constants. Your equation turns into one algebraic equation for Y, the transform of the answer.
Take y double prime plus y equals zero, starting at one with zero slope. Transforming gives s squared Y minus s plus Y equals zero. Solving leaves Y as s over s squared plus one, whose table line is cosine of t.
Split hard transforms with partial fractions into first order chunks. Constants over s minus a invert to exponentials, and shifted quadratics invert to sines, cosines, and their damped cousins. Each chunk inverts by sight from the table.
Forcing that switches on at time a wears the step u of t minus a, which is zero before a and one from a onward. Match the argument first, then the transform gains the factor e to minus a s. A mismatched argument is the classic slip, so align before you transform.
Transform with the starting values inside, solve the algebra, split into table pieces, and shift delayed forcing with a matched step.
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Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.