Order and degree of differential equations

Order is read straight off the equation: find the highest derivative that appears, and its order (first, second, ...) is the order of the equation.

where: $\dfrac{dy}{dx}$ is the first derivative of $y$ with respect to $x$, and $\dfrac{d^2y}{dx^2}$ is the second derivative; the order is simply the largest such index occurring anywhere in the equation.

Degree is stricter. It exists only when the equation, after clearing radicals and derivative-fractions, is a polynomial equation in its derivatives — no derivative may sit inside a square root, inside a fraction whose denominator contains a derivative, or inside a transcendental function such as $\sin$, $\cos$, $\log$ or $e^{(\cdot)}$. When that condition holds, the degree is the exponent on the highest-order derivative.

where: the highest-order derivative here is $\dfrac{d^2y}{dx^2}$, appearing to the power $2$; the lower-order term $\left(\dfrac{dy}{dx}\right)^3$ does not affect the degree at all — only the highest-order derivative's exponent counts. If the highest-order derivative is trapped under a radical or inside a non-polynomial function, and no valid rearrangement removes that trap, the degree is not defined — even though the order can still be stated with confidence. A frequent first move is to square both sides to clear a square root; this is legitimate only when doing so actually produces a genuine polynomial in the derivatives.

The definition of degree is deliberately tied to "polynomial in the derivatives" rather than to the equation's appearance, and the reason repays a moment's thought — it is why a square root sometimes matters and sometimes does not, and why the syllabus insists on checking before assigning a number.

Step 1. A polynomial in a variable $u$ has the form $a_n u^n + a_{n-1}u^{n-1} + \dots + a_0$, with only non-negative integer powers of $u$. The "degree" of that polynomial in $u$ is the largest exponent, $n$. Differential equations borrow this exact idea, treating the highest-order derivative as the variable $u$ whose degree we want.

Step 2. If a derivative sits inside $\sqrt{\ \cdot\ }$, inside $\sin(\cdot)$, $\log(\cdot)$ or $e^{(\cdot)}$, or in the denominator of a fraction, the expression is no longer a polynomial in that derivative — it has, in effect, a fractional, transcendental, or negative "power," none of which a polynomial's degree can describe. So the concept of degree simply has nothing to attach to, and it is left undefined rather than forced.

Step 3. Some radicals are removable: if only a lower-order derivative sits under the root, squaring both sides can produce an equation that genuinely is a polynomial once written out — in that case the trap was cosmetic, and after squaring the degree becomes well defined and is read off the resulting exponent of the highest-order derivative.

Step 4. Other radicals are not removable in this useful sense — if the highest-order derivative itself sits under a root together with other terms that will not clear by a single squaring, or if a transcendental function wraps a derivative, no amount of algebra turns the equation into a polynomial. The degree stays undefined, and the order alone still fully describes "how many times you'd differentiate" to reach that term.

Find the order and degree of $\dfrac{d^2y}{dx^2} = \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2}$.

Step 1. Locate every derivative present. The equation contains $\dfrac{d^2y}{dx^2}$, of order $2$, and $\dfrac{dy}{dx}$, of order $1$. The highest order occurring anywhere is $2$, so the order is $2$.

Step 2. Check which derivative sits under the radical. It is only $\dfrac{dy}{dx}$ — the lower-order derivative — that is trapped under the square root; the highest-order derivative $\dfrac{d^2y}{dx^2}$ stands outside it, already polynomial.

Step 3. Since only a lower-order term is under the root, squaring both sides is a valid, degree-preserving move: $\left(\dfrac{d^2y}{dx^2}\right)^2 = 1 + \left(\dfrac{dy}{dx}\right)^2.$ Both sides are now free of radicals, and every derivative appears with a non-negative integer power — the equation is a polynomial in its derivatives.

Step 4. Read the exponent on the highest-order derivative. $\dfrac{d^2y}{dx^2}$ is raised to the power $2$. The order is $2$ and the degree is $2$.

Competitive stretch

Contrast this with $\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + \sin\left(\dfrac{dy}{dx}\right) = 0$. The highest-order derivative is again $\dfrac{d^2y}{dx^2}$, so the order is still $2$. But here the trigonometric term $\sin\left(\dfrac{dy}{dx}\right)$ wraps a derivative in a transcendental function, and no algebraic step removes it. The equation is not a polynomial in its derivatives, so the degree is not defined, even though the order is a clean $2$. The two examples together show order and degree are genuinely independent readings of the same equation.

• Treating order and degree as the same question. Order asks "which derivative is highest" (a position question); degree asks "what power is that highest derivative raised to, once the equation is polynomial" (an exponent question). A low-order equation can have a high degree, and vice versa.
• Forcing a degree onto an equation that has none. If the highest-order derivative sits under a root, in a denominator, or inside $\sin$, $\log$ or $e^{(\cdot)}$, with no way to clear it, the degree is undefined — writing down a number anyway is the single most common CBSE mistake here.
• Squaring carelessly. Radicals must be cleared before the degree is read, but only clearing the derivative that was actually under the root is valid; squaring unrelated terms too, or losing track of which derivative is now highest, gives a wrong degree.
• Missing a lower-order derivative trapped in a transcendental function. Even if the highest-order derivative itself is perfectly polynomial, a lower-order derivative sitting inside $\sin$, $\cos$, $\log$, or similar still makes the whole equation fail the "polynomial in its derivatives" test, so the degree remains undefined.

Ask the tutor: if you are unsure whether an equation's degree is defined, ask your AI tutor to check with you whether it is a genuine polynomial in its derivatives before assigning a number.