The Laplace transform converts a function of t into a function of s. Under it, differentiation becomes multiplication.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The key property
The transform of y′ is s times the transform of y, minus the initial value. Derivatives become algebra, initial conditions included.
The strategy
Transform the whole equation, solve the resulting algebraic equation for the transform of y, then transform back.
Working from a table
The transform of 1 is 1/s, of t is 1/s², and of e^(at) is 1/(s − a). A short table covers most problems.
Initial conditions come free
They enter during the transform itself, so there is no separate step for finding constants at the end.
Where it shines
Discontinuous forcing, such as a switch flipping or a sudden impulse, is awkward otherwise and routine here.
Turning calculus into algebra
The Laplace transform converts a differential equation into an algebraic one in a new variable. Solve the algebra, then transform back. The difficulty moves from calculus to lookup and partial fractions.
Derivatives become multiplication
The transform of a derivative is s times the transform, minus the initial value. That is why initial conditions are built in from the start rather than fitted at the end.
Getting back is the work
Inverting usually means manipulating the expression into forms found in a table, often via partial fractions. That step, not the forward transform, is where the effort goes.
When it is worth using
It excels with discontinuous or impulsive forcing, which other methods handle awkwardly. For a smooth, simple equation the characteristic equation is usually faster.
Step 2: Try It Yourself
Tap and try it out.
- Point(1, 0.50)
Step 3: Watch an Example
One step at a time.
Watch Nils Transform an Equation
Nils applies the Laplace transform to y′ + 3y = 0 with y = 5 at t = 0.
- Step 1
He transforms the derivative, getting sY minus the initial value 5.
Step 4: Your Turn
Practice makes it stick.
The Table
Problem 1 of 2
The transform of e^(4t) is 1/(s − a). What is a?
The Inverse
Problem 2 of 2
Y = 7/(s + 2). The solution is Ce^(kt). What is C?
Into the s Domain
1 of 8
The transform of the constant 1 is 1 over s to what power?
2 of 8
The transform of t is 1 over s to what power?
3 of 8
The transform of e^(6t) is 1/(s − a). What is a?
4 of 8
Y = 3/(s + 5). The solution is Ce^(kt). What is k?
5 of 8
The transform of y′ is sY minus what quantity? Enter 1 for the initial value, 2 for zero.
6 of 8
Y = 9/(s − 1). The solution is Ce^(kt). What is C?
7 of 8
Order the steps of solving with the Laplace transform.
- 1Substitute the initial conditions as they appear
- 2Solve the algebraic equation for Y
- 3Take the inverse transform to recover y
- 4Transform the whole equation into the s domain
8 of 8
Under the Laplace transform, differentiation becomes which operation? Enter 1 multiplication, 2 addition.
Step 5: Quick Check
Show what you know.
Question 1 of 2
Y = 4/(s + 3). The solution is Ce^(kt). What is k?
Question 2 of 2
What does the Laplace transform do to differentiation?
What You Learned
- The Laplace transform sends a function of t to a function of s, turning derivatives into multiplication.
- Transform the equation, solve algebraically for Y, then invert to recover y.
- Initial conditions enter during the transform, so no constants remain to find at the end.