Solving a Differential Equation by the Transform: The Method
Chapter Seventy-Five
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 166 to 167 of 303
In one line
Transform every term, solve the algebra for Y(s), invert, and the initial conditions were used at the first step.
MU's label
"Solution of Ordinary Linear Differential Equations with Constant Coefficients". This is what the whole of Module 1's second half has been for.
The method, in five steps
- Transform every term of the equation, using the derivative theorem for each derivative and the table for the right-hand side.
- Substitute the initial values, which the derivative theorem has already put into the equation.
- Collect the terms in Y(s) and solve for Y, which is school algebra.
- Simplify F(s), usually by partial fractions or by completing the square.
- Invert to get y(t).
The two transform rows you need
L{diff(y(t), t)} = s Y(s) - y(0)
L{diff(y(t), (t, 2))} = s^2 Y(s) - s y(0) - diff(y(0), t)
Everything else comes off the ordinary table.
Worked, a first-order equation
Solve y prime plus 3y equals 0, with y(0) = 2.
Step one. Transforming: sY minus y(0), plus 3Y, equals 0.
Step two. y(0) is 2, so sY minus 2 plus 3Y equals 0.
Step three. Collecting: Y(s plus 3) equals 2, so Y equals 2 over (s plus 3).
Step four. Nothing to simplify.
Step five. Invert.
L⁻¹{2/(s + 3)} = 2 e^(-3t)
So y equals 2 e to the minus 3t. And the check, which is the same check the machine performs on every solution in this book: substitute it back.
dy/dx + 3y = 0
y = 2 e^(-3x)
Notice what did not happen. There was no general solution, no arbitrary constant, and no second stage of fitting the initial value. The 2 walked into the algebra at step two.
Worked, with a right-hand side
Solve y prime plus 2y equals e to the minus t, with y(0) = 0.
Transforming: sY minus 0, plus 2Y, equals one over (s plus 1).
Collecting: Y(s plus 2) equals one over (s plus 1), so Y equals one over (s plus 1)(s plus 2).
Partial fractions:
1/((s + 1)(s + 2)) = 1/(s + 1) - 1/(s + 2)
Inverting:
L⁻¹{1/((s + 1)(s + 2))} = e^(-t) - e^(-2t)
And the check:
dy/dx + 2y = e^(-x)
y = e^(-x) - e^(-2x)
The solution satisfies the equation, and at x = 0 it gives 1 minus 1, which is 0, matching the initial condition. Both halves of the problem are verified.
Worked, a second-order equation
Solve y double prime plus 4y equals 0, with y(0) = 1 and y prime of 0 = 0.
Transforming the second derivative: s squared Y minus s times 1 minus 0. So the equation is s squared Y minus s plus 4Y equals 0.
Solving a Differential Equation by the Transform: The Method
Collecting: Y(s squared plus 4) equals s, so Y equals s over (s squared plus 4).
L⁻¹{s/(s^2 + 4)} = cos(2t)
d2y/dx2 + 4y = 0
y = cos(2x)
Both initial conditions check: cos 0 is 1, and the derivative, minus 2 sin 2x, is 0 at x = 0.
Why the method is worth having
Set it against the methods of Module 2, which solve the same equations.
| Module 2 | The transform | |
|---|---|---|
| Complementary function | found separately | never appears |
| Particular integral | found separately | never appears |
| Arbitrary constants | found, then fitted | never appear |
| Initial conditions | used at the end | used at the start |
| Discontinuous input | solved in pieces | one extra factor |
The last row is where the method is not merely tidier but genuinely better, and the chapter on equations driven by a step or an impulse shows it.
The three places it goes wrong
Forgetting an initial value. The derivative theorem has a minus y(0) in it, and dropping it changes the answer completely. If your answer does not satisfy the initial condition, this is almost always why.
Getting the second derivative's terms wrong. It is s squared Y minus s y(0) minus y prime of 0, with the s on the y(0) and not on the y prime of 0. Swapping them is common.
Stopping at Y(s). The answer to a differential equation is a function of t. An answer left as a function of s is worth nothing, however correct.
Check yourself
dy/dx - y = 0
y = C e^x
dy/dx + y = 1
y = 1 + C e^(-x)
d2y/dx2 - y = 0
y = C1 e^x + C2 e^(-x)
Those three are written with arbitrary constants because they are checked as general solutions; a transform problem would fix the constants from the initial values. For the second, with y(0) = 0, the constant is minus 1 and the answer is 1 minus e to the minus x.
L⁻¹{1/(s(s + 1))} = 1 - e^(-t)
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.