A first-order ordinary differential equation in normal form is y′=F(x,y).y'=F(x,y). It is separable if the right-hand side is a product of a function of xx and a function of y.y.

y′=f(x) g(y)y'=f(x)\,g(y)

Where g(y)≠0g(y)\neq 0 the equation is y′g(y)=f(x),\frac{y'}{g(y)}=f(x), and by the chain rule and the second fundamental theorem of calculus a differentiable yy solves it if and only if

∫dyg(y)=∫f(x) dx,\int\frac{dy}{g(y)}=\int f(x)\,dx,

where the left-hand side is an antiderivative of 1g\frac{1}{g} evaluated at y(x),y(x), which defines yy implicitly up to a constant of integration. Each zero y0y_0 of gg gives the constant solution y=y0.y=y_0. If ff and gg are continuous and gg is C1C^1 then solutions through points with g(y)≠0g(y)\neq 0 are unique and never reach a zero of g.g. For example, y′=kyy'=ky has the solutions y=C ekx,y=C\,e^{kx}, which grow exponentially when k>0k>0 and decay exponentially when k<0.k<0.

A first-order equation is linear if it can be written with continuous coefficient functions pp and qq on an interval II as

y′+p(x) y=q(x).y'+p(x)\,y=q(x).

Let PP be an antiderivative of pp on I.I. The integrating factor μ(x):=eP(x)\mu(x):=e^{P(x)} satisfies μ′=p μ,\mu'=p\,\mu, so by the product rule the equation multiplied by μ\mu is (μ y)′=μ q,\left(\mu\,y\right)'=\mu\,q, and integrating gives the general solution on I.I.

y(x)=1μ(x)(∫μ(x) q(x) dx+C)=e−P(x)∫eP(x)q(x) dx+C e−P(x)y(x)=\frac{1}{\mu(x)}\left(\int\mu(x)\,q(x)\,dx+C\right)=e^{-P(x)}\int e^{P(x)}q(x)\,dx+C\,e^{-P(x)}

The solution is the sum of a particular solution and the general solution C e−P(x)C\,e^{-P(x)} of the homogeneous equation y′+p y=0.y'+p\,y=0. Every initial value problem y(x0)=y0y\left(x_0\right)=y_0 with x0∈Ix_0\in I has exactly one solution, which exists on all of I.I. For example, y′+1xy=1y'+\frac{1}{x}y=1 on (0,∞)\left(0,\infty\right) has μ=x\mu=x and general solution y=x2+Cx.y=\frac{x}{2}+\frac{C}{x}.

The Bernoulli equation y′+p(x) y=q(x) yny'+p(x)\,y=q(x)\,y^n with n≠0,1n\neq 0,1 is non-linear, but the substitution v=y1−nv=y^{1-n} turns it into the linear equation v′+(1−n)p(x) v=(1−n)q(x).v'+\left(1-n\right)p(x)\,v=\left(1-n\right)q(x). Similarly, a homogeneous equation of degree zero y′=F(yx)y'=F\left(\frac{y}{x}\right) becomes separable under the substitution v=yx,v=\frac{y}{x}, since x v′=F(v)−v.x\,v'=F(v)-v.

Let U⊆R2U\subseteq\mathbb{R}^2 be open and M,N:U→RM,N:U\rightarrow\mathbb{R} be C1C^1 with N≠0.N\neq 0. The equation y′=−M(x,y)N(x,y),y'=-\frac{M(x,y)}{N(x,y)}, written in differential form as

M(x,y) dx+N(x,y) dy=0,M(x,y)\,dx+N(x,y)\,dy=0,

is exact on UU if the vector field (M,N)\left(M,N\right) is conservative on U,U, that is, if there exists a potential ψ:U→R\psi:U\rightarrow\mathbb{R} with ψx=M\psi_x=M and ψy=N,\psi_y=N, where subscripts denote partial derivatives. By the chain rule the solutions are then the level sets of the potential,

ψ(x,y)=C,\psi\left(x,y\right)=C,

and each level set defines a solution y(x)y(x) implicitly wherever ψy≠0.\psi_y\neq 0. An exact equation satisfies the exactness condition My=NxM_y=N_x on UU by Schwarz’s theorem, and the converse holds when UU is simply connected, since then every irrotational field is conservative. The potential is found by integrating MM with respect to x,x, with a constant of integration that is a function of y,y, and fixing that function from ψy=N.\psi_y=N. For example, (2xy+1)dx+(x2+2y)dy=0\left(2xy+1\right)dx+\left(x^2+2y\right)dy=0 is exact with ψ=x2y+x+y2.\psi=x^2y+x+y^2.

An equation that is not exact can sometimes be made exact by multiplying it by an integrating factor μ(x,y)≠0,\mu(x,y)\neq 0, which must satisfy (μM)y=(μN)x.\left(\mu M\right)_y=\left(\mu N\right)_x. If My−NxN\frac{M_y-N_x}{N} depends on xx alone then μ(x)=exp⁡∫My−NxN dx\mu(x)=\exp\int\frac{M_y-N_x}{N}\,dx is an integrating factor, and if Nx−MyM\frac{N_x-M_y}{M} depends on yy alone then μ(y)=exp⁡∫Nx−MyM dy\mu(y)=\exp\int\frac{N_x-M_y}{M}\,dy is one. A linear equation (p y−q)dx+dy=0\left(p\,y-q\right)dx+dy=0 is the case My−NxN=p(x),\frac{M_y-N_x}{N}=p(x), which recovers its integrating factor eP(x).e^{P(x)}.

A first-order autonomous equation has a right-hand side that does not depend on x,x,

y′=f(y),y'=f(y),

with ff a C1C^1 function on an interval, and is separable. A zero y∗y^* of ff is an equilibrium point, also called a critical point, and the constant function y=y∗y=y^* is an equilibrium solution. By uniqueness no other solution reaches an equilibrium point, so between consecutive equilibrium points every solution is strictly monotonic, increasing where f>0f>0 and decreasing where f<0,f<0, and converges to an equilibrium point or diverges as x→∞.x\rightarrow\infty. The phase line records the equilibrium points and the sign of ff between them, which determines the direction of every solution.

An equilibrium point y∗y^* is stable if every solution that starts sufficiently close to y∗y^* stays close to it for all x≥x0,x\geq x_0,

∀ε>0 ∃δ>0:∣y(x0)−y∗∣<δ⇒∣y(x)−y∗∣<ε  ∀ x≥x0,\forall\varepsilon>0\,\exists\delta>0:\lvert y\left(x_0\right)-y^*\rvert<\delta\Rightarrow\lvert y(x)-y^*\rvert<\varepsilon\,\,\forall\,x\geq x_0,

asymptotically stable if it is stable and every solution that starts sufficiently close to y∗y^* converges to it as x→∞,x\rightarrow\infty, and unstable if it is not stable. On the phase line an equilibrium point is asymptotically stable if ff changes sign from positive to negative at it and unstable if ff changes sign from negative to positive. The linearisation of the equation at y∗y^* is u′=f′(y∗)uu'=f'\left(y^*\right)u for the deviation u=y−y∗,u=y-y^*, whose solutions are u=C ef′(y∗)x,u=C\,e^{f'\left(y^*\right)x}, and the linearisation criterion states that y∗y^* is asymptotically stable if f′(y∗)<0f'\left(y^*\right)<0 and unstable if f′(y∗)>0.f'\left(y^*\right)>0. If f′(y∗)=0f'\left(y^*\right)=0 the criterion is inconclusive: y′=−y3y'=-y^3 has an asymptotically stable equilibrium at 00 and y′=y3y'=y^3 an unstable one. A semi-stable equilibrium point, such as 00 for y′=y2,y'=y^2, attracts solutions from one side and repels them from the other.

The logistic equation models a quantity that grows at rate r>0r>0 when small and saturates at the carrying capacity K>0.K>0.

y′=r y(1−yK),y(x)=K1+(Ky0−1)e−rxy'=r\,y\left(1-\frac{y}{K}\right),\qquad y(x)=\frac{K}{1+\left(\frac{K}{y_0}-1\right)e^{-rx}}

Its equilibrium points are 0,0, which is unstable since f′(0)=r>0,f'(0)=r>0, and K,K, which is asymptotically stable since f′(K)=−r<0,f'(K)=-r<0, and every solution with y0>0y_0>0 converges to K.K.