Solving IVPs with Laplace Transforms

Main idea

Transform the whole initial value problem at once. Every derivative turns into $s$ times $Y(s)$ minus a starting value, so the equation becomes algebra you solve for $Y$ , and inverting $Y$ gives the answer with the initial conditions already in it.

\mathcal L\{y'\} = s{\htmlData{sym=0}{Y(s)}} - y(0), \qquad \mathcal L\{y''\} = s^2{\htmlData{sym=0}{Y(s)}} - sy(0) - y'(0)

What the symbols mean

$Y(s)$
Transform of the solution
The unknown, now found with algebra.

Explanation

An initial value problem (IVP) is simply a differential equation plus the starting conditions, like where a spring starts and how fast it's moving. Solving one with Laplace transforms means trading the calculus problem for an algebra problem, which is often much quicker. The rule $\mathcal L\{y''\} = s^2Y(s) - sy(0) - y'(0)$ looks like a mouthful, but it's just two simple parts: $s^2Y(s)$ (taking two derivatives becomes multiplying by $s$ twice) and $-sy(0) - y'(0)$ (the starting position and starting velocity, tagging along). Once every derivative is replaced like this, $Y$ is just an unknown in an equation you can solve with ordinary algebra. It's fewer moving parts than it looks.

This is the payoff. Back in the first lesson of this unit we promised that the transform turns derivatives into multiplication by $s$ . Here's why. Integrate by parts:

\mathcal L\{y'\} = \int_0^\infty e^{-st}y'\,dt = \Big[e^{-st}y\Big]_0^\infty + s\int_0^\infty e^{-st}y\,dt = -y(0) + sY(s).

The boundary term at infinity vanishes because $e^{-st}$ wins, and the one at $0$ leaves behind $y(0)$ . Apply the same rule to $y''$ , treating it as the derivative of $y'$ , and you get $s^2Y - sy(0) - y'(0)$ . Notice that the starting values show up on their own. You never have to plug them in at the end.

Steps:

Transform both sides, using the derivative rules.
Solve the resulting algebraic equation for $Y(s)$ .
Use partial fractions to write $Y$ as table entries.
Invert to get $y(t)$ .

Compare this with undetermined coefficients. There's no general solution, no separate guess for $y_p$ , and no solving for constants at the end: the initial conditions are used in step 1. The method really shines when the forcing switches on and off or hits all at once, which is where the next two lessons are going.

A common mistake. It's tempting to write $\mathcal L\{y''\} = s^2Y$ , because a derivative is "multiply by $s$ ". It isn't: each derivative also drops off a starting value, so $\mathcal L\{y''\} = s^2Y - sy(0) - y'(0)$ . Leaving those terms out is the same as assuming the system starts at rest, and you'd get the wrong answer whenever it doesn't.

See it

This is the charge on a capacitor in a circuit switched on at t = 0, with q″ + 2q′ + 5q = 5 and q(0) = q′(0) = 0. Transforming gives the algebra problem (s² + 2s + 5)Q = 5/s, and inverting gives q(t) = 1 − e^(−t)(cos 2t + ½ sin 2t). As you can see, the charge overshoots to about 1.21 C at t = π/2 ≈ 1.57, then wobbles and settles at 1 C.

Worked examples

Example 1

Solve $y'' - 3y' + 2y = e^{3t}$ , $y(0) = 1$ , $y'(0) = 0$ .

1
Transform both sides, with $y(0) = 1$ and $y'(0) = 0$ going into the derivative rules: $(s^2Y - s) - 3(sY - 1) + 2Y = \frac{1}{s - 3}$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $y' + 2y = 0$ , $y(0) = 3$ .

Problem 2+5 XP

Solve $y'' + y = 0$ , $y(0) = 0$ , $y'(0) = 1$ .

Problem 3+5 XP

Solve $y' - y = 1$ , $y(0) = 0$ .

Problem 4+5 XP

Solve $y'' - 3y' + 2y = e^{3t}$ , $y(0) = 1$ , $y'(0) = 0$ .

Problem 5+5 XP

Solve $y'' + 4y = 8$ , $y(0) = 0$ , $y'(0) = 0$ .

Problem 6+5 XP

A spring obeying $u'' + 9u = 0$ is released from $u(0) = 1$ with velocity $u'(0) = 3$ . Solve it with Laplace transforms.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $y'' + 2y' + 2y = 2$ with $y(0) = 0$ and $y'(0) = 1$ , using Laplace transforms.

Where it's used

Circuit simulators and control-system design tools solve linear ODEs this way, especially with switched or pulsed inputs.

Switching a Circuit On and OffDiff Eq · Electrical Engineering

Step Functions IVPs with Step Functions

Solving IVPs with Laplace Transforms

Main idea

\mathcal L\{y'\} = s{\htmlData{sym=0}{Y(s)}} - y(0), \qquad \mathcal L\{y''\} = s^2{\htmlData{sym=0}{Y(s)}} - sy(0) - y'(0)

What the symbols mean

$Y(s)$
Transform of the solution
The unknown, now found with algebra.

Explanation

This is the payoff. Back in the first lesson of this unit we promised that the transform turns derivatives into multiplication by $s$ . Here's why. Integrate by parts:

\mathcal L\{y'\} = \int_0^\infty e^{-st}y'\,dt = \Big[e^{-st}y\Big]_0^\infty + s\int_0^\infty e^{-st}y\,dt = -y(0) + sY(s).

Steps:

Transform both sides, using the derivative rules.
Solve the resulting algebraic equation for $Y(s)$ .
Use partial fractions to write $Y$ as table entries.
Invert to get $y(t)$ .

See it

Worked examples

Example 1

Solve $y'' - 3y' + 2y = e^{3t}$ , $y(0) = 1$ , $y'(0) = 0$ .

1
Transform both sides, with $y(0) = 1$ and $y'(0) = 0$ going into the derivative rules: $(s^2Y - s) - 3(sY - 1) + 2Y = \frac{1}{s - 3}$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $y' + 2y = 0$ , $y(0) = 3$ .

Problem 2+5 XP

Solve $y'' + y = 0$ , $y(0) = 0$ , $y'(0) = 1$ .

Problem 3+5 XP

Solve $y' - y = 1$ , $y(0) = 0$ .

Problem 4+5 XP

Solve $y'' - 3y' + 2y = e^{3t}$ , $y(0) = 1$ , $y'(0) = 0$ .

Problem 5+5 XP

Solve $y'' + 4y = 8$ , $y(0) = 0$ , $y'(0) = 0$ .

Problem 6+5 XP

A spring obeying $u'' + 9u = 0$ is released from $u(0) = 1$ with velocity $u'(0) = 3$ . Solve it with Laplace transforms.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $y'' + 2y' + 2y = 2$ with $y(0) = 0$ and $y'(0) = 1$ , using Laplace transforms.

Where it's used

Circuit simulators and control-system design tools solve linear ODEs this way, especially with switched or pulsed inputs.

Switching a Circuit On and OffDiff Eq · Electrical Engineering