First-order ODE classification

An ordinary differential equation (ODE) relates an unknown function $y(x)$ to its derivatives $y', y'', \dots$ Two labels are attached before anything else is attempted. The order is the order of the highest derivative appearing in the equation. The degree is the power to which that highest-order derivative is raised, once the equation has been cleared of fractions and radicals so that it is polynomial in the derivatives. Order and degree are independent counts — one about which derivative, the other about what power.

where: the highest derivative present is $dy/dx$ (order 1), and it is cubed (degree 3). By contrast $y'' + 3y' + 2y = 0$ has order 2 (because of $y''$) and degree 1 (that $y''$ appears only to the first power). An ODE is linear if the dependent variable $y$ and every derivative of it occur only to the first power, with no products such as $yy'$ and no nonlinear function of $y$ such as $\sin y$, $e^y$, or $y^2$. Any equation failing this test is nonlinear. Linearity does not care about $x$ — an equation can have $x^2$, $\sin x$, or $e^x$ multiplying $y$ or $y'$ and still be linear; it is only $y$ and its derivatives that must stay to the first power, unmixed. A first-order, first-degree equation $\dfrac{dy}{dx} = f(x,y)$ is called variables separable exactly when $f(x,y)$ factors as a product of a pure function of $x$ and a pure function of $y$, $f(x,y) = g(x)h(y)$. Dividing through by $h(y)$ isolates each variable on its own side:

where: $g(x)$ collects every $x$-dependent factor and $h(y)$ every $y$-dependent factor of the right-hand side; the split must be exact, not approximate. Integrating each side independently — one integral in $y$, one in $x$ — gives the general solution

where: $C$ is a single arbitrary constant of integration (the two integrals do not each need their own constant — one is enough, since a difference of constants is still just a constant). A first-order ODE's general solution is a one-parameter family of curves; supplying an initial condition $y(x_0)=y_0$ selects one curve from the family, the particular solution.

Working through these four checks in order — order, degree, linear or not, separable or not — is the practical habit the rest of the unit builds on. Misclassifying an equation almost always means reaching for a technique that cannot actually solve it.

Solve $\dfrac{dy}{dx} = \dfrac{x^2}{y^2}$, given $y(0) = 2$, and then extend it to a second initial condition.

Step 1 — classify. The equation is order 1 (only $dy/dx$ appears) and degree 1 (that derivative is to the first power). It is nonlinear because of the $y^2$ in the denominator, which is equivalent to $y^{-2}$ multiplying $dy/dx$ — not a first-power appearance of $y$. Writing the right side as $x^2 \cdot \frac{1}{y^2}$ shows it is separable, with $g(x) = x^2$ and $h(y) = y^2$.

Step 2 — separate. $y^2\,dy = x^2\,dx$.

Step 3 — integrate both sides. $\displaystyle\int y^2\,dy = \int x^2\,dx \implies \frac{y^3}{3} = \frac{x^3}{3} + C_1$. Multiplying through by 3 and renaming the constant gives $y^3 = x^3 + C$.

Step 4 — apply the initial condition. At $x=0$, $y=2$: $2^3 = 0^3 + C$, so $C = 8$.

Step 5 — write the particular solution. $y^3 = x^3 + 8$, so $y = (x^3+8)^{1/3}$. Checking by implicit differentiation, $3y^2y' = 3x^2$ gives $y' = x^2/y^2$, matching the original ODE, so the particular solution is $y = (x^3+8)^{1/3}$.

Competitive stretch

The same family $y^3 = x^3 + C$ answers a slightly harder question: for which initial points $(x_0, y_0)$ does the solution curve pass through the origin's neighbourhood without $y$ ever vanishing on the way? Since $y = (x^3+C)^{1/3}$ is defined and smooth for every real $x$ whenever $C \neq 0$ makes the cube root well-behaved (real cube roots exist for all real arguments, unlike square roots), the only delicate value is $C = 0$, i.e. $y_0^3 = x_0^3$, which forces $y = x$ — a solution curve that passes straight through the origin. Every other value of $C$ traces a curve that never touches $y=0$ except possibly where $x^3 = -C$, confirming that the constant from the initial condition genuinely controls the shape of the whole family, not just a single point on it.

• Confusing order with degree. Order counts which derivative is highest; degree is the power of that highest derivative once the equation is rationalized — they are independent numbers and neither determines the other.
• Forgetting the constant of integration until later. Add $C$ immediately after integrating both sides; delaying it risks losing or duplicating terms during rearrangement.
• Assuming every first-order ODE is separable. Only right-hand sides that factor as $g(x)h(y)$ split this way. $\dfrac{dy}{dx} = x+y$ is not separable — it is linear, and needs a different method covered later in the unit.
• Mislabelling nonlinear equations as linear. Watch for products like $yy'$, powers of $y$ above one, or $y$ trapped inside $\sin y$, $e^y$, or similar — any of these makes the equation nonlinear regardless of how the $x$-terms look.
• Losing a constant solution when dividing by $h(y)$. If $h(y_0) = 0$ for some constant $y_0$, then $y = y_0$ is itself a solution of the original ODE, but dividing by $h(y)$ during separation silently discards it — always check for such singular solutions separately.

Ask the tutor: unsure whether a tricky right-hand side actually factors as $g(x)h(y)$, or want another initial-value problem checked step by step? Ask the AI tutor to classify and solve it with you.