Order and degree

Open any differential equation and two numbers instantly tell you what kind of animal you are dealing with: its order — how many times it has been differentiated — and its degree — the power of the highest derivative once the equation is cleared of roots and fractions. These two labels are the first thing ISC Class 12 asks you to read off, before any solving method is even mentioned, because the Differential Equations section of the Calculus unit restricts solving to first-order, first-degree equations, while classifying equations of any order is fair game. Get order and degree right and you also unlock the twin ideas of general and particular solutions, and the reverse skill of forming a differential equation from a family of curves by eliminating its arbitrary constants. This article walks through the definitions, one grounded derivation of the "forming" process, and two fully worked classification examples — exactly the toolkit the syllabus expects before the actual solving techniques begin.

The core insight

The order of a differential equation is the order of its highest derivative, and its degree is that derivative's power once the equation is a polynomial in derivatives — undefined if it can never be made one.

The core idea

Order. Scan the equation for the highest-order derivative present — $\frac{dy}{dx}$ , $\frac{d^2y}{dx^2}$ , and so on — and its order is the order of the differential equation. There is no ambiguity here: order is purely about which derivative appears, not about any power attached to it.

Degree. Once the equation is rewritten as a polynomial in its derivatives (no radicals, no fractional or negative powers on any derivative), the degree is the positive-integer power carried by the highest-order derivative. If no amount of algebraic manipulation can turn the equation into such a polynomial — for instance because a derivative sits inside $\sin(\cdot)$ , $\log(\cdot)$ or $e^{(\cdot)}$ — the degree is not defined.

\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + y = 0 \quad\Rightarrow\quad \text{order } 2,\ \text{degree } 3

where: the highest derivative present is $\frac{d^2y}{dx^2}$ , fixing order $2$ , and it is already raised to a whole-number power with no radicals anywhere, so that power, $3$ , is read off directly as the degree.

General versus particular solutions. A general solution is one that carries as many independent arbitrary constants as the order of the equation; a particular solution fixes those constants using given initial or boundary conditions, collapsing the whole family to one curve.

Forming a differential equation. Run the logic backwards: start from a family of curves with $n$ arbitrary constants, differentiate the defining equation $n$ times, and eliminate the constants algebraically. What survives is a differential equation of order $n$ satisfied by every member of the family — the exact inverse of solving.

Figure

A flow diagram showing a family of curves with constants being differentiated repeatedly and the constants eliminated to leave a differential equation, with an arrow looping back labelled solving returns to the family

The forming direction: eliminating constants to reach an equation

Derivation AForming the differential equation of a two-constant family

The goal is to show, step by step and with a number attached at every stage, exactly how differentiating twice and eliminating two arbitrary constants from the curve family $y = A\cos x + B\sin x$ produces a specific order-2 differential equation that every member of the family obeys, which is the precise mechanism the syllabus calls "forming a differential equation" and the reverse of the solving process it restricts to first-order equations.

Step 1. Count the constants. The family $y = A\cos x + B\sin x$ carries two independent arbitrary constants, $A$ and $B$ , so the target differential equation must have order $2$ — differentiate exactly twice, once per constant.

Step 2. Differentiate once. $\dfrac{dy}{dx} = -A\sin x + B\cos x.$ Both constants are still present, so a second differentiation is needed before elimination is possible.

Step 3. Differentiate again. $\dfrac{d^2y}{dx^2} = -A\cos x - B\sin x.$ Notice the right-hand side is exactly $-(A\cos x + B\sin x)$ , which is just $-y$ from the original equation — the constants have not been isolated by brute force but have cancelled through recognition of the original expression.

Step 4. Substitute back to eliminate $A$ and $B$ entirely. $\dfrac{d^2y}{dx^2} = -y \;\Longrightarrow\; \dfrac{d^2y}{dx^2} + y = 0.$ No arbitrary constant appears anywhere in this equation, confirming the elimination is complete; every curve in the family, for any choice of $A$ and $B$ , satisfies it.

Step 5. Read off order and degree. The highest derivative present is $\dfrac{d^2y}{dx^2}$ , giving order 2 — matching the two constants eliminated, as it must. That derivative appears to the first power with no radicals, giving degree 1.

Worked example

State the order and degree of $\left(\dfrac{d^2y}{dx^2}\right)^2 + \sqrt{\dfrac{dy}{dx}} = x$ , then form the differential equation for the single-constant family $y = mx$ .

Step 1. Identify the order. The highest derivative in the given equation is $\dfrac{d^2y}{dx^2}$ , so the order is 2, regardless of what happens next with the radical.

Step 2. Clear the radical before touching degree. The term $\sqrt{\dfrac{dy}{dx}}$ blocks the equation from being a polynomial in derivatives, so isolate it and square both sides: $\dfrac{dy}{dx} = \left[x - \left(\dfrac{d^2y}{dx^2}\right)^2\right]^2.$ Every term is now polynomial in the derivatives.

Step 3. Read the degree from the rationalised form. The highest-order derivative, $\dfrac{d^2y}{dx^2}$ , appears with power $2$ on the right after expansion, so the degree is 2. (Had the equation instead trapped a derivative inside $\sin(\cdot)$ or a similar transcendental function with no algebraic escape, the degree would be not defined rather than any specific number.)

Step 4. Form the differential equation for $y = mx$ (lines through the origin, $m$ arbitrary). One constant means differentiate once: $\dfrac{dy}{dx} = m$ . Substitute back into $y = mx$ to eliminate $m$ : $y = \left(\dfrac{dy}{dx}\right)x$ , so $x\dfrac{dy}{dx} - y = 0.$

Final answer: the given equation has order 2, degree 2; and $x\dfrac{dy}{dx} - y = 0$ is the order-1, degree-1 equation satisfied by every line through the origin.

Competitive stretch

If instead the family had three arbitrary constants — say $y = A + Bx + Ce^x$ — the same logic demands differentiating three times before all three constants cancel, producing an order-3 equation. The rule scales directly: the number of differentiations, and hence the order of the resulting equation, always equals the number of independent arbitrary constants in the original family, never more and never fewer.

Common traps

Reading off degree before clearing radicals or fractional powers — the equation must first be rationalised into a polynomial in derivatives; degree read from the un-rationalised form is meaningless.
Confusing order with degree — order asks which derivative is highest; degree asks the power of that highest derivative, a completely separate question.
Assuming a degree always exists — a derivative trapped inside a transcendental function that cannot be algebraically freed makes the degree not defined, not "1" by default.
Differentiating the wrong number of times when forming an equation — the number of differentiations must exactly match the number of independent arbitrary constants; too few leaves constants behind, too many overshoots the required order.

Ask the tutor: Stuck on whether a tricky equation's degree is defined, or how many times to differentiate a family with several constants? Ask the AI tutor to walk through your exact equation step by step.

Why does clearing radicals matter so much for degree but not for order?

Order only cares about which derivative is the highest one present, and that identification does not change whether the equation is written with a radical or not — the highest derivative stays the highest derivative either way. Degree, however, is defined as a power, and a radical or fractional exponent on a derivative is not a well-defined integer power. So you must first rationalise the equation into a polynomial in its derivatives, and only then can you honestly read off the power attached to the highest-order term as the degree.

If a differential equation's degree is not defined, does that mean it has no solution?

No — "degree not defined" is purely a classification statement about how the equation is written, not a statement about whether solutions exist. Equations like those involving $\sin(y')$ or $e^{y''}$ can absolutely have solutions; they simply fall outside the classification scheme built on treating the equation as a polynomial in derivatives. At the ISC level you are only asked to recognise and state that the degree is not defined for such equations, not to solve them.

When forming a differential equation, how do I know when the constants are all correctly eliminated?

Check that no arbitrary constant appears anywhere in your final equation — it should contain only $x$ , $y$ , and derivatives of $y$ . A useful habit, as in the $y = A\cos x + B\sin x$ example, is to look for the original expression reappearing inside a later derivative (here $\frac{d^2y}{dx^2} = -y$ ), which lets you substitute directly instead of solving simultaneous equations for $A$ and $B$ . If a constant still lingers after differentiating as many times as there are constants, recheck your algebra before assuming you need another differentiation.