# Linearity of DE's

• I
Hey all,

I don't understand what makes a differential equation (DE) linear.
I found this: "x y' = 1 is non-linear because y' is not multiplied by a constant"
but then also this: "x' + (t^2)x = 0 is linear in x".

t^2 also isn't a constant.
So why is this equation linear?

Last edited:

Related Differential Equations News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
A differential operator ##\hat L## is linear if ##\hat L (a f_1 + b f_2) = a\hat L f_1 + b \hat L f_2##, where a and b are constants and the fs functions.

A differential equation for a function ##f## is linear if it can be written as ##\hat L f = g##, where g is some fixed function that does not depend on f. If g=0 the differential equation is homogeneous.

x y’’(x) = 1 is an inhomogeneous linear differential equation. Your second example is a homogeneous linear differential equation.

A differential operator ##\hat L## is linear if ##\hat L (a f_1 + b f_2) = a\hat L f_1 + b \hat L f_2##, where a and b are constants and the fs functions.

A differential equation for a function ##f## is linear if it can be written as ##\hat L f = g##, where g is some fixed function that does not depend on f. If g=0 the differential equation is homogeneous.

x y’’(x) = 1 is an inhomogeneous linear differential equation. Your second example is a homogeneous linear differential equation.
I don't understand, still. What makes y(t) different from t2. They're both non-constants.
If you define y(t)=t^2 then you would be able to do y(t) (a+b) = ay(t)+by(t) = at^2 +bt^2, right?

Last edited:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
You do not "define" ##y(t)##, you need to solve the differential equation to find out what ##y(t)## is. It has nothing to do with whether or not the coefficient in front of ##y## is constant in the independent variable or not. The only thing that matters is the property I described above, i.e., whether or not the differential equation can be written ##\hat L y = g## for some linear differential operator ##\hat L## or not. To take your first example, your differential operator would be ##\hat L = x (d/dx)## and ##g = 1##, leading to
$$\hat L y(x) = x\frac{dy}{dx} = x y'(x) = 1.$$
The operator ##\hat L## is linear because
$$\hat L [a_1 y_1(x) + a_2 y_2(x)] = x \frac{d}{dx}[a_1 y_1(x) + a_2 y_2(x)] = a_1 x \frac{dy_1}{dx} + a_2 x \frac{dy_2}{dx} = a_1 \hat L y_1 + a_2 \hat L y_2.$$

Edit: Effectively, the DE is linear if it is a sum of terms where each term contains only one factor of ##y(x)## or its derivatives. It does not matter whether the coefficients are constant in the independent variable or not.

Edit 2: So the general linear ODE of order ##n## is of the form
$$\sum_{k = 0}^n f_k(x) \frac{d^ky}{dx^k} = g(x).$$
The ODE is homogeneous if ##g(x) = 0## and otherwise inhomogeneous.

• AVBs2Systems
You do not "define" ##y(t)##, you need to solve the differential equation to find out what ##y(t)## is. It has nothing to do with whether or not the coefficient in front of ##y## is constant in the independent variable or not. The only thing that matters is the property I described above, i.e., whether or not the differential equation can be written ##\hat L y = g## for some linear differential operator ##\hat L## or not. To take your first example, your differential operator would be ##\hat L = x (d/dx)## and ##g = 1##, leading to
$$\hat L y(x) = x\frac{dy}{dx} = x y'(x) = 1.$$
The operator ##\hat L## is linear because
$$\hat L [a_1 y_1(x) + a_2 y_2(x)] = x \frac{d}{dx}[a_1 y_1(x) + a_2 y_2(x)] = a_1 x \frac{dy_1}{dx} + a_2 x \frac{dy_2}{dx} = a_1 \hat L y_1 + a_2 \hat L y_2.$$

Edit: Effectively, the DE is linear if it is a sum of terms where each term contains only one factor of ##y(x)## or its derivatives. It does not matter whether the coefficients are constant in the independent variable or not.

Edit 2: So the general linear ODE of order ##n## is of the form
$$\sum_{k = 0}^n f_k(x) \frac{d^ky}{dx^k} = g(x).$$
The ODE is homogeneous if ##g(x) = 0## and otherwise inhomogeneous.
Can you show me an example of a non-linear equation and can you show me (using the operator) why it isn't linear, please?

Last edited:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
An example of a non-linear differential equation would be
$$y'(x) + y(x)^2 = 0.$$
It is non-linear because
$$[y_1(x)+y_2(x)]' + [y_1(x)+y_2(x)]^2 \neq [y_1'(x) + y_1(x)^2] + [y_2'(x) + y_2(x)^2].$$

An example of a non-linear differential equation would be
$$y'(x) + y(x)^2 = 0.$$
It is non-linear because
$$[y_1(x)+y_2(x)]' + [y_1(x)+y_2(x)]^2 \neq [y_1'(x) + y_1(x)^2] + [y_2'(x) + y_2(x)^2].$$

Why am I so confused...?

Last edited:
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Your differential operator cannot depend on the dependent function itself. Furthermore, you cannot write
$$y'(x) + y(x)^2 = y(x) [d/dx + y(x)],$$
the derivative needs to act on ##y(x)##. You could write it as
$$[d/dx + y(x)]y(x),$$
but it is still not linear because the first parenthesis depends on ##y(x)##.

Your differential operator cannot depend on the dependent function itself. Furthermore, you cannot write
$$y'(x) + y(x)^2 = y(x) [d/dx + y(x)],$$
the derivative needs to act on ##y(x)##. You could write it as
$$[d/dx + y(x)]y(x),$$
but it is still not linear because the first parenthesis depends on ##y(x)##.
I am sorry. I don't understand any of this.

Hello,

First of all my apologies. Turns out I never had the subject of operators yet. After watching a video with an explanation I think it is now totally clear on what a linear operator is defined as and how to apply an operator to a function.

I see now what you were saying in the beginning. I might've been doing alternative math all this time, which is never a good idea because it's fictional.

Thanks for the help ;-)

-Yael

Last edited by a moderator: