Order, Degree and What Counts as a Solution

A differential equation relates a function to its derivatives. Two labels classify every DE on the paper:

Order = the highest derivative present. Degree = the power of that highest derivative after the equation is polynomial in all derivatives. If the derivatives sit inside a sine, exponential, log or root that cannot be cleared, the degree is not defined — a favourite one-marker.

$$\left(\frac{d^2y}{dx^2}\right)^3 + \left(\frac{dy}{dx}\right)^2 + y = 0 \quad\Rightarrow\quad \text{order } 2,\ \text{degree } 3$$ $$\frac{dy}{dx} + \sin\left(\frac{dy}{dx}\right) = 0 \quad\Rightarrow\quad \text{order } 1,\ \text{degree not defined}$$

General vs particular solution: the general solution of an $n$th-order DE carries exactly $n$ arbitrary constants; a particular solution has none — the constants are fixed by the given conditions. Verifying a solution means differentiating the candidate and substituting back, never solving from scratch.

Method 1 — Variables Separable

If the equation can be pushed into the shape $g(y)\,dy = f(x)\,dx$, integrate both sides and add one constant: $$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \;\Rightarrow\; \int\frac{dy}{1+y^2} = \int\frac{dx}{1+x^2} \;\Rightarrow\; \tan^{-1}y = \tan^{-1}x + C$$ Leave implicit answers implicit — do not fight to isolate $y$ unless the question demands it.

The Method-Picking Drill

Board questions never announce the method. Run this check in order:

1. Can $x$-terms and $y$-terms be separated? Separate and integrate.
2. Is $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ (every term of the same total degree)? It is homogeneous — substitute $y = vx$.
3. Does it fit $\frac{dy}{dx} + P(x)\,y = Q(x)$? It is linear — build the integrating factor.
4. None of the above in $y$? Try swapping roles: the equation may be linear in $x$ with $\frac{dx}{dy} + P(y)\,x = Q(y)$.

One equation, one method — the drill picks it in under a minute.

Worked Examples — Order, Degree, Verification, Separation

Example 1 — Order and degree

State the order and degree of $\left(\dfrac{d^2y}{dx^2}\right)^3 + \left(\dfrac{dy}{dx}\right)^2 + y = 0$ and of $\dfrac{dy}{dx} + \sin\left(\dfrac{dy}{dx}\right) = 0$.

Step 1 — first equation: highest derivative is $\frac{d^2y}{dx^2}$ (order $2$), raised to power $3$ in a polynomial equation — degree $3$.

Step 2 — second equation: order $1$, but $\frac{dy}{dx}$ sits inside a sine, so no polynomial form exists.

Answer: order $2$, degree $3$; and order $1$, degree not defined.

Example 2 — Verifying a solution

Verify that $y = A\cos x + B\sin x$ is a solution of $y'' + y = 0$.

Step 1 — differentiate twice: $y' = -A\sin x + B\cos x$, $y'' = -A\cos x - B\sin x = -y$.

Step 2 — substitute: $y'' + y = -y + y = 0$.

Answer: verified — with two arbitrary constants for a second-order equation, this is the general solution.

Example 3 — Separation of variables

Solve $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$.

Step 1 — separate: $\dfrac{dy}{1 + y^2} = \dfrac{dx}{1 + x^2}$.

Step 2 — integrate both sides: $\tan^{-1}y = \tan^{-1}x + C$.

Answer: $\tan^{-1}y = \tan^{-1}x + C$ — one constant, on one side, and the implicit form is a complete answer.

Example 4 — Separation with exponentials

Solve $\dfrac{dy}{dx} = e^{x+y}$.

Step 1 — split the exponential: $e^{x+y} = e^x e^y$, so $e^{-y}\,dy = e^x\,dx$.

Step 2 — integrate: $-e^{-y} = e^x + C_1$.

Answer: $e^x + e^{-y} = C$ — tidy the constant's sign at the end; examiners accept any equivalent constant.

Example 5 — A particular solution

Solve $\dfrac{dy}{dx} = y\tan x$ given $y(0) = 1$.

Step 1 — separate and integrate: $\int\dfrac{dy}{y} = \int\tan x\,dx$ gives $\log\vert y\vert = \log\vert\sec x\vert + \log C$, so $y = C\sec x$.

Step 2 — apply the condition: $1 = C\sec 0 = C$.

Answer: $y = \sec x$ — a particular solution has no arbitrary constant left.

Example 6 — Writing the constant as $\log C$

Solve $x\dfrac{dy}{dx} = y$, $x > 0$, $y > 0$.

Step 1 — separate: $\dfrac{dy}{y} = \dfrac{dx}{x}$, so $\log y = \log x + \log C$.

Step 2 — combine the logs: $\log y = \log(Cx)$.

Answer: $y = Cx$ — when both sides produce logs, writing the constant as $\log C$ collapses the answer in one line.

Method 2 — Homogeneous Equations

A first-order DE is homogeneous when it can be written as $\frac{dy}{dx} = F\left(\frac{y}{x}\right)$ — equivalently, replacing $x$ by $\lambda x$ and $y$ by $\lambda y$ leaves the right side unchanged.

The fixed routine: substitute $y = vx$, so that $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ The equation always becomes separable in $v$ and $x$. Solve, then replace $v = \frac{y}{x}$ at the end. (If the equation is instead $\frac{dx}{dy} = G\left(\frac{x}{y}\right)$, use $x = vy$ symmetrically.)

Method 3 — Linear Differential Equations

The standard form is $$\frac{dy}{dx} + P(x)\,y = Q(x)$$ First force the equation into this shape (divide out any coefficient of $\frac{dy}{dx}$), then read off $P$ and $Q$. The integrating factor is $$\text{IF} = e^{\int P\,dx}$$ and the solution is always $$y\cdot\text{IF} = \int Q\cdot\text{IF}\,dx + C$$ Useful simplification: $e^{a\log u} = u^a$, so $P = \frac{2}{x}$ gives IF $= x^2$, and $P = -3\cot x$ gives IF $= \frac{1}{\sin^3 x}$ — never leave an IF as an unevaluated exponential of a log.

Linear in $x$: when the equation resists all $y$-forms, rewrite it as $$\frac{dx}{dy} + P(y)\,x = Q(y), \qquad \text{IF} = e^{\int P\,dy}, \qquad x\cdot\text{IF} = \int Q\cdot\text{IF}\,dy + C$$ The roles of $x$ and $y$ swap completely — everything else is identical.

Worked Examples — Homogeneous and Linear

Example 7 — Homogeneous, the base case

Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$.

Step 1 — confirm homogeneity and substitute $y = vx$: $v + x\dfrac{dv}{dx} = 1 + v$.

Step 2 — separate: $x\dfrac{dv}{dx} = 1$, so $dv = \dfrac{dx}{x}$ and $v = \log\vert x\vert + C$.

Step 3 — restore $v = \frac{y}{x}$:

Answer: $y = x\log\vert x\vert + Cx$.

Example 8 — Homogeneous with a trig twist

Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \tan\dfrac{y}{x}$.

Step 1 — substitute $y = vx$: $v + x\dfrac{dv}{dx} = v + \tan v$, so $x\dfrac{dv}{dx} = \tan v$.

Step 2 — separate: $\int\cot v\,dv = \int\dfrac{dx}{x}$ gives $\log\vert\sin v\vert = \log\vert x\vert + \log C$.

Answer: $\sin\dfrac{y}{x} = Cx$ — the whole solution is two integrals from the table.

Example 9 — Homogeneous with an initial condition

Solve $\dfrac{dy}{dx} = \dfrac{2xy + y^2}{x^2}$ with $y(1) = 1$.

Step 1 — substitute $y = vx$: $v + x\dfrac{dv}{dx} = 2v + v^2$, so $x\dfrac{dv}{dx} = v + v^2$.

Step 2 — separate with partial fractions: $\dfrac{dv}{v(1 + v)} = \dfrac{dx}{x}$ gives $\log\left\vert\dfrac{v}{1+v}\right\vert = \log\vert x\vert + \log C$, so $\dfrac{y}{x + y} = Cx$.

Step 3 — apply $y(1) = 1$: $\dfrac{1}{2} = C$.

Answer: $2y = x(x + y)$, that is $y = \dfrac{x^2}{2 - x}$ — partial fractions from the Integrals chapter finish homogeneous problems constantly.

Example 10 — Linear, the template

Solve $\dfrac{dy}{dx} + y = e^x$.

Step 1 — read off $P = 1$, $Q = e^x$: IF $= e^{\int 1\,dx} = e^x$.

Step 2 — the solution formula: $y e^x = \int e^x\cdot e^x\,dx = \dfrac{e^{2x}}{2} + C$.

Answer: $y = \dfrac{e^x}{2} + Ce^{-x}$.

Example 11 — Linear with IF a power of $x$

Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2$.

Step 1 — IF: $e^{\int dx/x} = e^{\log x} = x$.

Step 2 — solve: $yx = \int x^2\cdot x\,dx = \dfrac{x^4}{4} + C$.

Answer: $y = \dfrac{x^3}{4} + \dfrac{C}{x}$ — always simplify $e^{\log(\cdot)}$ before integrating.

Example 12 — Linear with a condition and a trig IF

Solve $\dfrac{dy}{dx} - 3y\cot x = \sin 2x$ given $y\left(\dfrac{\pi}{2}\right) = 2$.

Step 1 — IF: $e^{-3\int\cot x\,dx} = e^{-3\log\vert\sin x\vert} = \dfrac{1}{\sin^3 x}$.

Step 2 — solve: $\dfrac{y}{\sin^3 x} = \int\dfrac{2\sin x\cos x}{\sin^3 x}\,dx = \int 2\cot x\csc x\,dx = -\dfrac{2}{\sin x} + C$.

Step 3 — apply the condition: at $x = \dfrac{\pi}{2}$: $2 = -2 + C$, so $C = 4$.

Answer: $y = 4\sin^3 x - 2\sin^2 x$.

Example 13 — Linear in $x$

Solve $y\dfrac{dx}{dy} = x + 2y^2$.

Step 1 — standard form in $x$: $\dfrac{dx}{dy} - \dfrac{x}{y} = 2y$, so $P(y) = -\dfrac{1}{y}$, $Q(y) = 2y$.

Step 2 — IF: $e^{-\int dy/y} = \dfrac{1}{y}$.

Step 3 — solve: $\dfrac{x}{y} = \int 2y\cdot\dfrac{1}{y}\,dy = 2y + C$.

Answer: $x = 2y^2 + Cy$ — when the $y$-form looks hopeless, the $x$-form is usually a three-line linear.