Differential Equations • Topic 1 of 3

Differential Equations: Order, Degree & Formation

A differential equation (DE) is an equation that connects an unknown function with one or more of its derivatives. Where an algebraic equation asks "which numbers satisfy this?", a differential equation asks "which functions satisfy this?". Examples: $\dfrac{dy}{dx}=2x$, $\dfrac{d^2y}{dx^2}+y=0$, and $\left(\dfrac{dy}{dx}\right)^2+y=\sin x$. Because the unknown here is a function, the answer to a DE is a function (or a whole family of them), not a single value.

An equation involving derivatives of a function of a single independent variable is an ordinary differential equation — the only kind in the Class 12 syllabus. (Equations with partial derivatives of a multivariable function are called partial differential equations and are studied later.)

Order of a differential equation

The order is the order of the highest derivative that actually appears in the equation. So $\dfrac{dy}{dx}+y=0$ has order $1$; $\dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}+2y=0$ has order $2$; and $\dfrac{d^3y}{dx^3}+y^2=x$ has order $3$. The order is always a positive whole number, and it tells you how many times the function has been differentiated.

Degree of a differential equation

The degree is the power (exponent) of the highest-order derivative, taken after the equation has been written as a polynomial in all of its derivatives — that is, after clearing every radical and fractional power of the derivatives. The degree is defined only when such a polynomial form exists.

For example $\left(\dfrac{d^2y}{dx^2}\right)^3+\dfrac{dy}{dx}=0$ has order $2$ and degree $3$. But the degree is not defined for equations where a derivative sits inside a transcendental function or cannot be freed of radicals — for instance $\dfrac{d^2y}{dx^2}+\sin\!\left(\dfrac{dy}{dx}\right)=0$ has no degree, because $\sin\!\left(\dfrac{dy}{dx}\right)$ can never be turned into a polynomial in $\dfrac{dy}{dx}$.

If the equation contains a radical involving derivatives, first isolate the radical and square (or raise to the appropriate power) to remove it, then read off the degree. For $\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\dfrac{d^2y}{dx^2}$, squaring gives $1+\left(\dfrac{dy}{dx}\right)^2=\left(\dfrac{d^2y}{dx^2}\right)^2$; the highest derivative is order $2$ raised to power $2$, so order $2$, degree $2$.

EquationOrderDegree
$\dfrac{dy}{dx}+y=x$$1$$1$
$\left(\dfrac{dy}{dx}\right)^3+2y=0$$1$$3$
$\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^4=0$$2$$1$
$\left(\dfrac{d^2y}{dx^2}\right)^2+\left(\dfrac{dy}{dx}\right)^3=0$$2$$2$
$\dfrac{d^2y}{dx^2}+\sin\!\left(\dfrac{dy}{dx}\right)=0$$2$not defined

General and particular solutions

A function (with or without arbitrary constants) that satisfies a DE is a solution. The general solution contains as many independent arbitrary constants as the order of the equation: a first-order DE has one, a second-order DE has two. Fixing those constants from given conditions (such as $y(0)=1$) produces a particular solution — one specific member of the family.

For example, $y=Ce^{x}$ is the general solution of $\dfrac{dy}{dx}=y$; applying $y(0)=3$ forces $C=3$, giving the particular solution $y=3e^{x}$.

Verifying that a function is a solution

To check whether a given function solves a DE, compute the required derivatives and substitute them into the equation. If the two sides become identically equal (for all $x$), the function is a solution; otherwise it is not. This is a substitution exercise — no integration is needed.

Formation of a differential equation

A family of curves described by an equation with $n$ independent arbitrary constants gives rise to a differential equation of order $n$. The recipe is:

  • Count the arbitrary constants — say there are $n$ of them.
  • Differentiate the given equation $n$ times with respect to $x$.
  • Use the original equation together with the $n$ derived equations to eliminate all $n$ constants.
  • The relation that survives, free of every arbitrary constant, is the required differential equation. It is satisfied by the whole family at once.

So one differentiation removes one constant, two differentiations remove two, and so on. A family with one parameter (like $y=mx$) yields a first-order DE; a two-parameter family (like $y=mx+c$, or $y=Ae^{x}+Be^{-x}$) yields a second-order DE.

Deeper Insight — why order, degree and formation are really the same idea seen from two sides. Forming a DE destroys information (the constants), while solving a DE recovers it. Differentiating a one-parameter family once removes its single constant and produces a first-order equation; conversely, integrating a first-order equation once reintroduces exactly one constant. This is why the number of arbitrary constants in the general solution always equals the order — the two operations are inverses. Keep degree separate from order in your mind: order counts how many times you differentiated, degree measures the highest power of that top derivative once the equation is polynomial. Many marks are lost by quoting a degree for an equation that has none — always check for radicals or trig/exp/log wrapped around a derivative first.

A one-parameter family of straight lines y = c x through the origin, the general solution of x dy/dx = y General solution = a family of curves (one constant c) x y O c=2 c=1 c=½ c=−½ c=−1 One arbitrary constant c → order 1 (DE: x dy/dx = y)
1
Worked Example
Find the order and degree of $\dfrac{d^2y}{dx^2}+3\left(\dfrac{dy}{dx}\right)^2+y=0$.
Solution
  1. Highest derivative present is $\dfrac{d^2y}{dx^2}$, so the order is $2$.
  2. That highest derivative appears to the first power, so the degree is $1$ (the square is on the lower-order $\dfrac{dy}{dx}$, which does not affect degree).

Answer: Order $2$, degree $1$.

2
Worked Example
Find the order and degree of $\left(\dfrac{dy}{dx}\right)^3+2y=x$.
Solution
  1. The only derivative is $\dfrac{dy}{dx}$, which is first order.
  2. Its highest power is $3$, and the equation is already polynomial in the derivative.

Answer: Order $1$, degree $3$.

3
Worked Example
Form the differential equation of the family $y=mx$ ($m$ arbitrary).
Solution
  1. There is one arbitrary constant $m$, so differentiate once: $\dfrac{dy}{dx}=m$.
  2. Eliminate $m$ using $m=\dfrac{y}{x}$ from the original equation: $\dfrac{dy}{dx}=\dfrac{y}{x}$.

Answer: $x\dfrac{dy}{dx}=y$ (first order, as expected for one constant).

4
Worked Example
Verify that $y=e^{2x}$ is a solution of $\dfrac{dy}{dx}=2y$.
Solution
  1. Differentiate: $\dfrac{dy}{dx}=2e^{2x}$.
  2. Compute $2y=2e^{2x}$.
  3. Both sides agree for all $x$.

Answer: Yes; since $\dfrac{dy}{dx}=2e^{2x}=2y$, the function satisfies the equation.

5
Worked Example
Find the order and degree of $\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}=\dfrac{d^2y}{dx^2}$.
Solution
  1. Square both sides to clear the radical: $1+\left(\dfrac{dy}{dx}\right)^2=\left(\dfrac{d^2y}{dx^2}\right)^2$.
  2. The highest derivative is $\dfrac{d^2y}{dx^2}$ (order $2$), now raised to power $2$.

Answer: Order $2$, degree $2$.

6
Worked Example
State the order and degree (if defined) of $\dfrac{d^2y}{dx^2}+\cos\!\left(\dfrac{dy}{dx}\right)=0$.
Solution
  1. Highest derivative is second order, so order $=2$.
  2. The term $\cos\!\left(\dfrac{dy}{dx}\right)$ cannot be expressed as a polynomial in the derivatives, so no polynomial form exists.

Answer: Order $2$; degree not defined.

7
Worked Example
Verify that $y=A\cos x+B\sin x$ is a solution of $\dfrac{d^2y}{dx^2}+y=0$.
Solution
  1. $\dfrac{dy}{dx}=-A\sin x+B\cos x$.
  2. $\dfrac{d^2y}{dx^2}=-A\cos x-B\sin x=-y$.
  3. Hence $\dfrac{d^2y}{dx^2}+y=-y+y=0$.

Answer: Verified — the equation holds for all $x$, so $y=A\cos x+B\sin x$ is a solution.

8
Worked Example
Form the differential equation of the family of straight lines $y=mx+c$ ($m,c$ arbitrary).
Solution
  1. There are two arbitrary constants, so differentiate twice.
  2. First: $\dfrac{dy}{dx}=m$.
  3. Second: $\dfrac{d^2y}{dx^2}=0$ — both constants are now eliminated.

Answer: $\dfrac{d^2y}{dx^2}=0$ (second order, matching two constants).

9
Worked Example
Form the differential equation of the family $y=Ae^{x}+Be^{-x}$.
Solution
  1. Two constants $A,B$ — differentiate twice.
  2. $\dfrac{dy}{dx}=Ae^{x}-Be^{-x}$.
  3. $\dfrac{d^2y}{dx^2}=Ae^{x}+Be^{-x}=y$.

Answer: $\dfrac{d^2y}{dx^2}-y=0$.

10
Worked Example
Form the differential equation of the family of circles $x^2+y^2=a^2$ (centre at origin, radius $a$ arbitrary).
Solution
  1. One arbitrary constant $a$ — differentiate once.
  2. $2x+2y\dfrac{dy}{dx}=0$.
  3. Divide by $2$: the constant $a$ has already vanished.

Answer: $x+y\dfrac{dy}{dx}=0$.

11
Worked Example
Form the differential equation of the family $y=ae^{3x}+be^{-2x}$.
Solution
  1. Two constants — differentiate twice. $\dfrac{dy}{dx}=3ae^{3x}-2be^{-2x}$ and $\dfrac{d^2y}{dx^2}=9ae^{3x}+4be^{-2x}$.
  2. Note $y'+2y=5ae^{3x}$ and $y''+2y'=15ae^{3x}=3(y'+2y)$.
  3. Hence $y''+2y'=3y'+6y$, i.e. $y''-y'-6y=0$.

Answer: $\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-6y=0$.

12
Worked Example
Verify that $y=\sqrt{1+x^2}$ is a solution of $(1+x^2)\dfrac{dy}{dx}=xy$.
Solution
  1. $\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1+x^2}}$.
  2. LHS $=(1+x^2)\cdot\dfrac{x}{\sqrt{1+x^2}}=x\sqrt{1+x^2}$.
  3. RHS $=xy=x\sqrt{1+x^2}$.
  4. LHS $=$ RHS for all $x$.

Answer: Verified; $y=\sqrt{1+x^2}$ satisfies $(1+x^2)y'=xy$.

Key Points

  • A differential equation links a function with its derivatives; its solution is a function, not a number.
  • Order = order of the highest derivative present (a positive integer).
  • Degree = power of the highest-order derivative once the equation is polynomial in its derivatives; clear radicals first.
  • Degree is not defined when a derivative sits inside $\sin,\cos,\ln,e^{\,\cdot}$ etc. or cannot be made polynomial.
  • The general solution of an order-$n$ DE carries exactly $n$ arbitrary constants; fixing them via given conditions gives a particular solution.
  • Verify a solution by substituting the function and its derivatives back into the equation — no integration needed.
  • To form a DE from a family with $n$ constants: differentiate $n$ times and eliminate all $n$ constants.
  • Forming a DE and solving one are inverse processes — this is why the number of constants equals the order.
Tap an option to check your answer0 / 4
Q1.The degree of $\left(\dfrac{d^2y}{dx^2}\right)^3+\dfrac{dy}{dx}=0$ is:
Explanation: Highest-order derivative (2nd) is raised to power $3$.
Q2.The order of $\dfrac{dy}{dx}+y=x$ is:
Explanation: Highest derivative is first order.
Q3.The DE of the family $y=mx$ is:
Explanation: Eliminate $m$: $dy/dx=y/x$.
Q4.A general solution of an order-$n$ DE has how many arbitrary constants?
Explanation: Exactly $n$ constants.