Open the app

Solving IVPs with Laplace Transforms

Differential Equations · Section 6.2 · generated practice set

Transform the IVP, split with partial fractions or complete the square, and read the answer off the table.

Practice this set → Fresh numbers on every attempt. No account needed.

Method

The Laplace transform turns a differential equation into an algebra problem. Three steps, and the middle one is where the work is.

1. Transform

The derivative rules carry the initial conditions into the equation, which is why this method handles IVPs so directly:

$$mathcal{L}{y'} = sY - y(0), qquad mathcal{L}{y''} = s^2Y - s,y(0) - y'(0).$$

For $ay'' + by' + cy = g(t)$ this gives $$left(as^2 + bs + c ight)Y(s) = a,s,y(0) + a,y'(0) + b,y(0) + G(s).$$ The polynomial multiplying $Y$ is the characteristic polynomial with $r$ renamed $s$ — every single time.

2. Solve for $Y(s)$, then invert

Which technique depends on the denominator:

  • Distinct real roots → partial fractions. The cover-up method is fastest: to get the coefficient over $(s-a)$, delete that factor and evaluate what remains at $s = a$.
  • Irreducible quadratic (negative discriminant) → complete the square, do not force partial fractions. Rewrite as $(s-alpha)^2 + eta^2$ and split the numerator to match the two table entries below.
  • Repeated roots → partial fractions with both $dfrac{A}{s-a}$ and $dfrac{B}{(s-a)^2}$.

3. Read off $y(t)$

$$mathcal{L}^{-1}left{ rac{s-alpha}{(s-alpha)^2+eta^2} ight} = e^{alpha t}coseta t, qquad mathcal{L}^{-1}left{ rac{eta}{(s-alpha)^2+eta^2} ight} = e^{alpha t}sineta t.$$

Note the numerators: the cosine entry needs $s - alpha$ (shifted), the sine entry needs the bare $eta$. Matching them is what the numerator-splitting in step 2 is for.

Definitions and theorems

Transform of derivatives
$mathcal{L}{y'} = sY(s) - y(0)$ and $mathcal{L}{y''} = s^2Y(s) - sy(0) - y'(0)$, provided $y$ is of exponential order.
Cover-up rule
For a simple pole, the coefficient of $dfrac{1}{s-a}$ in $dfrac{N(s)}{(s-a)Q(s)}$ is $dfrac{N(a)}{Q(a)}$.
Shifted table entries
$mathcal{L}^{-1}left{dfrac{s-alpha}{(s-alpha)^2+eta^2} ight} = e^{alpha t}coseta t$ and $mathcal{L}^{-1}left{dfrac{eta}{(s-alpha)^2+eta^2} ight} = e^{alpha t}sineta t$.

Worked example

Solve $y'' + 4y' + 13y = 0$, $y(0) = 1$, $y'(0) = 2$ by Laplace transform.

  1. Transform: $left(s^2Y - s - 2 ight) + 4left(sY - 1 ight) + 13Y = 0$.
  2. Collect: $(s^2 + 4s + 13)Y = s + 6$. The constant is $y'(0) + 4y(0) = 2 + 4 = 6$.
  3. The discriminant is $16 - 52 < 0$, so complete the square: $s^2 + 4s + 13 = (s+2)^2 + 9$, giving $eta = 3$.
  4. Split the numerator to match: $s + 6 = (s+2) + 4$, so $Y = dfrac{s+2}{(s+2)^2+9} + dfrac43cdotdfrac{3}{(s+2)^2+9}$.
  5. $y(t) = e^{-2t}left(cos 3t + frac43sin 3t ight)$.

Common mistakes

  1. Dropping the $b,y(0)$ term. Transforming $by'$ produces $b(sY - y(0))$; the $-by(0)$ moves to the right side and joins the numerator. It is the most commonly lost term in the whole method.
  2. Forcing partial fractions on an irreducible quadratic. If the discriminant is negative there are no real linear factors. Complete the square.
  3. Numerator mismatch. $dfrac{1}{(s+2)^2+9}$ is not $e^{-2t}sin 3t$ — it is $ frac13 e^{-2t}sin 3t$. The sine entry needs $eta$ on top.
  4. Sign of the shift. $(s+2)^2 + 9$ corresponds to $e^{-2t}$, not $e^{2t}$.
  5. Applying initial conditions at the end. With Laplace they go in at the transform step; there is no $c_1, c_2$ to solve for afterwards.

Practice it

Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.

Open 6.2 →

More sets in Differential Equations

1.1Introduction to ODEs & Direction Fields1.2Solutions to Some ODEs1.3Classification of Differential Equations2.1First-Order Linear Equations2.2Separable Equations2.3Modeling with First-Order ODEs