Order, degree and what a solution is

What you will be able to do

• say what a differential equation is, and find its order and its degree;
• check by substitution that a given function is a solution;
• tell a general solution from a particular solution, and fix the constants from conditions.

A cup of tea

A cup of tea cools. The hotter it is, the faster it cools. So the rate of change of the temperature depends on the temperature itself. A rule like this does not give the temperature at a time. It gives a link between the temperature and its rate of change. Such a link is a differential equation.

A differential equation is an equation that contains a function of one variable and one or more of its derivatives. Here the function is $y$ and the variable is $x$. This chapter treats ordinary differential equations, where there is only one independent variable.

Order

The order is the order of the highest derivative in the equation. The derivative $\dfrac{dy}{dx}$ has order $1$, and $\dfrac{d^2y}{dx^2}$ has order $2$.

• $\dfrac{dy}{dx} + y = x$ has order $1$.
• $\dfrac{d^2y}{dx^2} + 3\dfrac{dy}{dx} + 2y = 0$ has order $2$.
• $\dfrac{d^3y}{dx^3} + \dfrac{dy}{dx} = e^x$ has order $3$.

Degree

The degree is the power of the highest-order derivative, after the equation has been written as a polynomial in its derivatives. Follow three steps.

1. Clear any root or fraction that contains a derivative.
2. Find the highest-order derivative.
3. Read its power. That power is the degree.

The degree is defined only when the equation is a polynomial in the derivatives. If a derivative sits inside $\sin$, $\cos$, $e^{(\ )}$ or $\ln$, the degree is not defined.

• $\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^4 + y = 0$ has order $2$ and degree $3$. The highest derivative is the second, and its power is $3$.
• $\dfrac{d^2y}{dx^2} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^3}$ has order $2$. Squaring both sides gives $\left(\dfrac{d^2y}{dx^2}\right)^2 = 1 + \left(\dfrac{dy}{dx}\right)^3$, so the degree is $2$.
• $\sin\dfrac{dy}{dx} = x$ has order $1$, and the degree is not defined.
• $\dfrac{d^2y}{dx^2} + e^{dy/dx} = 0$ has order $2$, and the degree is not defined, because a derivative is in the exponent.

Order and degree are both positive whole numbers, and the order does not depend on the degree. The degree is a term of the Indian syllabus. A-Level, IB and AP speak of the order only, and they say a first-order or second-order equation.

Solutions

A solution is a function $y = f(x)$ that makes the equation true for every $x$ in an interval. To check a solution, put the function and its derivatives into the equation.

Example. Is $y = e^{3x}$ a solution of $y'' - 5y' + 6y = 0$? Here $y' = 3e^{3x}$ and $y'' = 9e^{3x}$. Then $9e^{3x} - 15e^{3x} + 6e^{3x} = 0$. Yes, it is.

A first-order equation such as $\dfrac{dy}{dx} = 2x$ has a whole family of solutions, $y = x^2 + C$. The letter $C$ is an arbitrary constant. The family is the general solution. In general, an equation of order $n$ has a general solution with $n$ independent arbitrary constants.

A particular solution is one curve of the family. A condition picks it. For $\dfrac{dy}{dx} = 2x$ with the condition $y = 3$ when $x = 1$, we get $3 = 1 + C$, so $C = 2$ and the particular solution is $y = x^2 + 2$. A condition of this kind, given at one point, is called an initial condition. In IB it is an initial value problem.

More constants, more conditions

The equation $y'' - y' - 2y = 0$ has order $2$. Its general solution is $y = Ae^{2x} + Be^{-x}$, with two constants. Two conditions fix them, for example $y(0)$ and $y'(0)$. Move the sliders. The curve changes, and the equation stays true.

Worked example

The function $y = Ae^{2x} + Be^{-x}$ has $y(0) = 5$ and $y'(0) = 1$. Find $A$ and $B$.

1. Write the conditions. $y(0) = A + B = 5$. Next, $y' = 2Ae^{2x} - Be^{-x}$, so $y'(0) = 2A - B = 1$.
2. Add the two equations. The $B$ cancels: $3A = 6$, so $A = 2$.
3. Find $B$. $B = 5 - 2 = 3$.
4. The particular solution is $y = 2e^{2x} + 3e^{-x}$.

Practise with new numbers

Try it in Python

The equation is $y'' - 2y' + y = 0$. The program estimates $y'$ and $y''$ from nearby points, and prints $y'' - 2y' + y$ for four candidates. A solution gives zero.

Common mistakes

• Mixing up order and degree. The order counts derivatives. The degree is the power of the highest one.
• Reading the degree before clearing roots. $y'' = \sqrt{1 + y'}$ has degree $2$ after squaring, not $1$.
• Counting a degree when it is not defined. A derivative inside $\sin$ or $e^{(\ )}$ means no degree.
• Too few constants. The general solution of an order $n$ equation needs $n$ constants.
• Not substituting to check. Put the function and its derivatives into the equation. It takes one minute.

Remember these

• The order is the order of the highest derivative. The degree is its power, when the equation is a polynomial in the derivatives.
• A solution makes the equation true. The general solution has as many constants as the order.
• A particular solution comes from conditions, one for each constant.
• Check a solution by substituting it.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. State the order and the degree of $\left(\dfrac{d^3y}{dx^3}\right)^2 + 4\dfrac{d^2y}{dx^2} + y = x$.

2. State the order and the degree of $\sin\dfrac{dy}{dx} = x$.

3. Is $y = 3x^2$ a solution of $x\dfrac{dy}{dx} = 2y$?

4. How many arbitrary constants does the general solution of $y''' - y'' = 0$ have?

5. The equation $y'' = y$ has the solution $y = Pe^x + Qe^{-x}$. Find the particular solution with $y(0) = 2$ and $y'(0) = 0$.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.1 Basics of Differential Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on basic concepts and on general and particular solutions.