Lecture notes · Differential Equations · Lecture 13

Finding Particular Solutions to Inhomogeneous ODEs

MIT 18.03 Differential Equations, lecture 13, explained in plain English

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Part 1 · 0:00–10:00

This part explains why exponential functions make finding particular solutions to these equations much easier, and how a simple substitution rule does the work.

Key ideas

A particular solution, called yp, is one piece of the general solution. The general solution is yp plus the complementary solution, which solves the equation with the right side set to zero. 1:04
The inputs that matter most are exponentials of the form e to the alpha x. If alpha is allowed to be a complex number, this one form covers decaying exponentials, pure sine and cosine waves, and decaying oscillations. 3:45
The left side of the equation can be written as a polynomial operator p(D) acting on y. Here D means 'take a derivative,' so p(D) turns y into a combination of y, its first derivative, and its second derivative. 5:38
The substitution rule says that applying p(D) to e to the alpha x gives p(alpha) times e to the alpha x. In other words, you just replace every D with alpha. 7:19
The rule works because the derivative of e to the alpha x is alpha times e to the alpha x. Linearity, meaning derivatives of sums and scaled terms add up nicely, extends this from single derivatives to the whole polynomial. 8:43

Words to know

complex number: A number with a real part and an imaginary part, written like a + i times b. Here alpha can be one, which is what lets one exponential form cover sine and cosine too.
operator: A rule that takes in a function and gives back a new function. Here p(D) is a rule built from derivatives and constant numbers.
linearity: The property that the derivative of a sum is the sum of the derivatives, and constants can be pulled out. This lets results for single derivatives be added together.
complementary solution: The general solution to the same equation when the right side is set to zero. It is added to the particular solution to get the full answer.

Example

Think of pushing a swing with a steady, repeating rhythm. The push is like a pure vibration input, and the swing responds in a way you can predict from that rhythm. The lesson says a spring system can be given the same kind of steady vibration, and we want to see how it responds.

3 checked practice questions for this part are in the app.

Part 2 · 10:00–20:00

This part explains a shortcut for finding one specific solution to a differential equation when the input is an exponential, and shows how to handle sine waves using complex numbers.

Key ideas

When the input is an exponential, e to the alpha x, a particular solution is simply e to the alpha x divided by p of alpha. Here p of alpha means you plug alpha into the polynomial built from the equation's coefficients. 12:09
The lesson verifies the formula by substituting it back into the equation. Applying the polynomial to e to the alpha x multiplies it by p of alpha, and that cancels the division, leaving e to the alpha x. 13:46
The proof assumes p of alpha is not zero, because dividing by zero is not allowed. The lesson says the zero case is a separate, important case that it will come back to later. 15:04
A sine or cosine input can be handled by writing it as the imaginary part of a complex exponential. You solve the complex version, then take the imaginary part of that answer to get the real solution. 18:12
The constant 10 in the input simply carries through to the answer. This is because the operator is linear, so constants pass straight through it. 19:11

Words to know

particular solution: One specific answer that fits the whole equation, including the input on the right side. It is different from the general family of answers.
exponential input theorem: The lesson's name for the rule that says what the answer looks like when the input is an exponential like e to the alpha x.
complex exponential: An exponential whose power is a complex number, which includes an imaginary part with the letter i. It can be used to represent sine and cosine waves.
homogeneous equation: The same differential equation with the input on the right side set to zero. Its solutions form the base part of the general answer.

Example

Think of a swing pushed by a steady rhythm. The push is the input, and the swing's motion is the answer. Once you know the rhythm matches a simple pattern, you can predict the swing's motion right away, instead of working out every bump from scratch.

3 checked practice questions for this part are in the app.

Part 3 · 20:00–30:00

This part shows how to tidy up a messy complex answer to get a real particular solution, and introduces a shortcut rule for applying a polynomial of D to an exponential.

Key ideas

The original problem's right-hand side is the imaginary part of a complex version, so the answer you want is the imaginary part of the complex solution. 20:00
A messy complex fraction is easier to use if you multiply the top and bottom by the complex conjugate, which makes the bottom a plain real number. Then you can split the expression into real and imaginary parts. 22:12
A cosine plus a sine with equal coefficients can be rewritten as one cosine with a square root of two in front and a shift of pi over four. 24:33
When p of alpha is zero, the usual method fails, so the lecturer needs a new rule for that case. 24:58
The exponential shift rule lets you move an exponential past a polynomial in D, and when you do, D changes to D plus a. This avoids taking many derivatives by hand. 27:32

Words to know

imaginary part: The part of a complex number that is multiplied by i. A complex number has a real part and an imaginary part, like 3 + 2i.
complex conjugate: The same complex number with the sign of its imaginary part flipped. For example, the conjugate of 1 - i is 1 + i.
operator: A rule that takes in a function and gives back a new function, such as taking a derivative. Here, a polynomial in D is an operator.
exponential shift rule: A shortcut that says how to apply a polynomial in D to an exponential times a function, by changing D into D plus a.

Example

Think of a shift like moving a sign on a shop door. If the shop name is D, and you move the whole sign one step to the left, the name you read changes to D plus a. The rule tells you how the name changes when the exponential gets moved past it, so you don't have to rebuild the sign from scratch each time.

3 checked practice questions for this part are in the app.

Part 4 · 30:00–40:00

This part shows a shortcut for handling e to the ax inside a differential operator, and uses it to find particular solutions when the simple formula breaks down.

Key ideas

When a differential operator p(D) acts on e to the ax times u, you can pull e to the ax out to the front. Doing so changes every D into D plus a. 31:15
Once the rule works for D, it works for D squared, and then for any polynomial in D, because a polynomial is just a sum of powers of D with constant numbers in front. 31:34
To handle D squared, you don't start from scratch. You apply the rule for D twice, one step at a time. The same pattern continues to any power, which is proved by mathematical induction. 32:12
If p(a) equals zero, the simple formula fails. Instead, the particular solution is x times e to the ax, divided by the derivative of p at a. If a is a double root, the solution uses x squared times e to the ax, divided by the second derivative of p at a. 36:34

Words to know

Product rule: A rule for taking the derivative of two things multiplied together. You take the derivative of the first, times the second, then add the first times the derivative of the second.
Linearity: A rule that says you can split a sum apart and handle each piece separately, and constant numbers can be pulled out in front.
Mathematical induction: A proof method. You show a pattern works for the first case, then show that if it works for one case it also works for the next one. The lecturer uses it to extend the rule to any power of D.
Root (zero) of a polynomial: A number that makes the polynomial equal zero. A simple root appears once. A double root appears twice, which is possible here because the polynomial is quadratic.

Example

Think of a set of gears where turning one gear shifts the speed of the next by a fixed amount. Passing e to the ax through D is like shifting the speed by a, so every D becomes D plus a. The lecture does the same kind of shift to find a solution.

3 checked practice questions for this part are in the app.

Part 5 · 40:00–47:47

This part proves a shortcut formula for finding one particular solution to an ODE when the input is an exponential and the root is simple, then checks it on an example.

Key ideas

A particular solution is one answer that fits the equation with the given input. You only need one, not every possible answer. 42:57
If the root a is simple, the polynomial has a factor (D minus a) and does not have a repeated root at a. That means the derivative p prime at a is not zero, so it is safe to divide by it. 41:00
The proposed answer is x times e to the ax, divided by p prime of a. The lecturer checks it by applying the polynomial to it and seeing that the result is just e to the ax. 42:39
The exponential shift rule lets you move e to the ax out past the operator, and changing D to D plus a makes the calculation simple. The result reduces to a minus b, which cancels the same value in p prime of a. 43:34
In the example, the root 1 is simple. The particular solution is x times e to the x, divided by p prime of 1, which is negative one. So the answer is minus x e to the x. 47:20

Words to know

simple root: A solution to the polynomial equation that appears only once, not repeated. Here it means the polynomial has a factor (D minus a) that appears just one time.
particular solution: One specific answer to the full equation, including the input on the right side. It is different from the general answer that has no input.
exponential shift rule: A rule that lets you move an e to the ax term past the operator. You replace D with D plus a, so the calculation becomes simpler.
derivative: The rate of change of a function. p prime is the derivative of the polynomial, and p prime of a means that derivative evaluated at the point a.

Example

Think of a ball rolling into a bowl, with a steady push applied to it. The particular solution is one specific path the ball takes because of that push, not every possible path. Here, the lecturer finds one path that fits the push, and that is enough for the homework.

3 checked practice questions for this part are in the app.

← Lecture 10: Continuation: Complex Characteristic Roots · Lecture 15: Introduction to Fourier Series →

Notes written by AI (Claude) from the lecture's official transcript, so check anything important against the lecture. MIT OpenCourseWare · CC BY-NC-SA 4.0 · not affiliated with or endorsed by MIT. These notes are shared under the same license.

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