Basic concepts of differential equations

A differential equation is an equation involving an independent variable ($x$), a dependent variable ($y$), and at least one derivative of $y$ with respect to $x$. In full generality, an equation in $y = y(x)$ can be written as

where: $F$ stands for "some function of," $x$ is the independent variable, $y$ is the unknown dependent variable, and $\dfrac{d^ky}{dx^k}$ is the $k$-th derivative of $y$ with respect to $x$. Any equation fitting this pattern — from $\dfrac{dy}{dx} = x^2 + 3$ to far messier expressions — is a differential equation; an equation with no derivative in it, however complicated, is not. A solution of a differential equation is a function $y = f(x)$ (or a relation connecting $x$ and $y$) that satisfies the equation identically, for every $x$ in some interval, once it and its derivatives are substituted in. Verifying a proposed solution never requires solving anything new: differentiate the candidate function as many times as the equation demands, substitute every piece back in, and check the two sides match. Differential equations are classified along two axes:

Order is the order of the highest derivative present. An equation containing only $\dfrac{dy}{dx}$ is first order; one containing $\dfrac{d^2y}{dx^2}$ is second order, regardless of what lower-order terms also appear.

Degree is the power of the highest-order derivative, once the equation has been cleared of fractions and radicals in the derivatives so that it is a polynomial in $\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}, \ldots$. Degree is defined only when the equation can be written as such a polynomial; if the highest derivative sits inside a sine, log, or a fractional power that cannot be cleared, the degree is not defined.

The goal is to take a concrete differential equation and pin down both its order and its degree correctly, since mixing the two up (or reading degree off the wrong term) is the single most common slip when this topic is first tested, and the definitions above only become solid once applied to an actual equation rather than left as abstract statements.

Step 1. Write down the equation and scan every term for derivatives of $y$. Take

The terms present are $\left(\dfrac{d^2y}{dx^2}\right)^3$, $\dfrac{dy}{dx}$, and the algebraic term $x$.

Step 2. Identify the order by finding the highest derivative appearing anywhere in the equation. Here the derivatives present are $\dfrac{d^2y}{dx^2}$ (second order) and $\dfrac{dy}{dx}$ (first order). The highest is $\dfrac{d^2y}{dx^2}$, so the order is 2. Note that the order does not care about powers — only about how many times $y$ has been differentiated.

Step 3. Confirm the equation is already a polynomial in its derivatives (no derivative sits inside a root, a fraction with a derivative in the denominator, a sine, or a logarithm). It is: both derivative terms appear as plain powers added together. Only now is degree defined at all.

Step 4. Read the power attached to the highest-order derivative — not the highest power appearing anywhere in the equation. The highest-order derivative is $\dfrac{d^2y}{dx^2}$, and it is raised to the power $3$. So the degree is 3, even though $\dfrac{dy}{dx}$ appears only to the first power and no other term is cubed. Order 2, degree 3 fully classifies this equation.

Verify that $y = e^{2x}$ is a solution of the differential equation $\dfrac{dy}{dx} - 2y = 0$, and state its order and degree.

Step 1. Differentiate the given function. $y = e^{2x} \implies \dfrac{dy}{dx} = 2e^{2x}$.

Step 2. Substitute $y$ and $\dfrac{dy}{dx}$ into the left-hand side of the equation. $\dfrac{dy}{dx} - 2y = 2e^{2x} - 2\left(e^{2x}\right) = 2e^{2x} - 2e^{2x} = 0$.

Step 3. Compare with the right-hand side. The left-hand side equals $0$ for every value of $x$, matching the right-hand side identically — not just at one point.

Step 4. Classify the equation. Only $\dfrac{dy}{dx}$ appears, to the first power, so the equation is order $1$, degree $1$. $y = e^{2x}$ is confirmed as a solution of $\dfrac{dy}{dx} - 2y = 0$, an order-1, degree-1 equation.

Competitive stretch

Show that $y = A\cos x + B\sin x$, with $A$ and $B$ arbitrary constants, is a solution of $\dfrac{d^2y}{dx^2} + y = 0$.

Step 1. Find the first derivative. $\dfrac{dy}{dx} = -A\sin x + B\cos x$.

Step 2. Find the second derivative. $\dfrac{d^2y}{dx^2} = -A\cos x - B\sin x$.

Step 3. Substitute into the equation. $\dfrac{d^2y}{dx^2} + y = (-A\cos x - B\sin x) + (A\cos x + B\sin x) = 0$, satisfied identically. This is an order-2, degree-1 equation, and the presence of two arbitrary constants ($A$ and $B$) matches its order — a pattern the next topics build on. (Second-order solving is outside the CBSE syllabus; only verification is being demonstrated.)

• Confusing an algebraic equation with a differential equation. $x^2 + y^2 = 4$ contains no derivative and is not a differential equation on its own.
• Forgetting to substitute the derivative, not just the function. Verifying a solution needs both $y$ and every required derivative substituted back — plugging in $y$ alone proves nothing.
• Reading degree off the wrong term. Degree comes from the power of the highest-order derivative only, never from a lower-order derivative that happens to carry a bigger power.
• Assuming degree always exists. If the highest derivative is trapped inside a sine, a log, or a radical that cannot be cleared, the degree is simply undefined — do not force a number.

Ask the tutor: if you want to see order and degree worked out for a trickier equation, or want another function verified as a solution before moving to general and particular solutions, ask your AI tutor to walk through more examples with you.