First-Order Equations

A first-order ordinary differential equation in normal form is an equation $y'=F(x,y).$

Let $y'=F(x,y)$ be a first-order equation. The equation is separable if and only if $F$ is a product of a function of $x$ and a function of $y.$

$$y'=f(x)\,g(y)$$

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

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

where the left-hand side is an antiderivative of $\frac{1}{g}$ evaluated at $y(x).$ This defines $y$ implicitly up to a constant of integration. Separable equations have the following properties:

• each zero $y_0$ of $g$ gives the constant solution $y=y_0;$
• if $f$ and $g$ are continuous and $g$ is $C^1$ then solutions through points with $g(y)\neq 0$ are unique and never reach a zero of $g.$

Let $k\in\mathbb{R}.$ The equation $y'=ky$ is separable with solutions $y=C\,e^{kx}.$ The solutions grow exponentially when $k>0$ and decay exponentially when $k<0.$

Let $I$ be an interval and $p,q:I\rightarrow\mathbb{R}$ be continuous. A first-order linear equation is an equation of the form

$$y'+p(x)\,y=q(x).$$

Let $y'+p(x)\,y=q(x)$ be a first-order linear equation on $I$ and $P$ be an antiderivative of $p$ on $I.$ The integrating factor of the equation is the function $\mu(x):=e^{P(x)},$ which satisfies $\mu'=p\,\mu.$ By the product rule the equation multiplied by $\mu$ is $\left(\mu\,y\right)'=\mu\,q,$ and integrating gives the general solution on $I.$

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

First-order linear equations have the following properties:

• the general solution is the sum of a particular solution and the general solution $C\,e^{-P(x)}$ of the homogeneous equation $y'+p\,y=0;$
• every initial value problem $y\left(x_0\right)=y_0$ with $x_0\in I$ has exactly one solution, which exists on all of $I.$

For example, $y'+\frac{1}{x}y=1$ on $\left(0,\infty\right)$ has $\mu=x$ and general solution $y=\frac{x}{2}+\frac{C}{x}.$

Let $p,q$ be continuous on an interval and $n\neq 0,1.$ The Bernoulli equation is the non-linear equation $y'+p(x)\,y=q(x)\,y^n.$ The substitution $v=y^{1-n}$ turns the Bernoulli equation into the linear equation $v'+\left(1-n\right)p(x)\,v=\left(1-n\right)q(x).$

Let $F$ be a function of one variable. A homogeneous equation of degree zero is an equation of the form $y'=F\left(\frac{y}{x}\right).$ The substitution $v=\frac{y}{x}$ turns it into the separable equation $x\,v'=F(v)-v.$

Let $U\subseteq\mathbb{R}^2$ be open and $M,N:U\rightarrow\mathbb{R}$ be $C^1$ with $N\neq 0.$ The differential form of the equation $y'=-\frac{M(x,y)}{N(x,y)}$ is

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

Let $M(x,y)\,dx+N(x,y)\,dy=0$ be an equation in differential form on $U.$ The equation is exact on $U$ if and only if the vector field $\left(M,N\right)$ is conservative on $U,$ that is, if and only if there exists a potential $\psi:U\rightarrow\mathbb{R}$ with $\psi_x=M$ and $\psi_y=N,$ where subscripts denote partial derivatives. By the chain rule the solutions of an exact equation are the level sets of the potential,

$$\psi\left(x,y\right)=C,$$

and each level set defines a solution $y(x)$ implicitly wherever $\psi_y\neq 0.$

Let $M(x,y)\,dx+N(x,y)\,dy=0$ be an equation in differential form on $U.$ The exactness condition is the equation $M_y=N_x$ on $U.$ Exact equations have the following properties:

• an exact equation satisfies the exactness condition, by Schwarz’s theorem;
• if $U$ is simply connected then an equation satisfying the exactness condition is exact, since every irrotational field on a simply connected set is conservative;
• the potential is found by integrating $M$ with respect to $x,$ with a constant of integration that is a function of $y,$ and fixing that function from $\psi_y=N.$

For example, $\left(2xy+1\right)dx+\left(x^2+2y\right)dy=0$ is exact with $\psi=x^2y+x+y^2.$

Let $M(x,y)\,dx+N(x,y)\,dy=0$ be an equation in differential form that is not exact. An integrating factor of the equation is a function $\mu(x,y)\neq 0$ such that the equation multiplied by $\mu$ is exact, which requires $\left(\mu M\right)_y=\left(\mu N\right)_x.$ Integrating factors have the following properties:

• if $\frac{M_y-N_x}{N}$ depends on $x$ alone then $\mu(x)=\exp\int\frac{M_y-N_x}{N}\,dx$ is an integrating factor;
• if $\frac{N_x-M_y}{M}$ depends on $y$ alone then $\mu(y)=\exp\int\frac{N_x-M_y}{M}\,dy$ is an integrating factor;
• a linear equation written as $\left(p\,y-q\right)dx+dy=0$ has $\frac{M_y-N_x}{N}=p(x),$ which recovers its integrating factor $e^{P(x)}.$

Let $f$ be a $C^1$ function on an interval. A first-order autonomous equation is an equation whose right-hand side does not depend on $x.$

$$y'=f(y)$$

A first-order autonomous equation is separable.

Let $y'=f(y)$ be a first-order autonomous equation. An equilibrium point or critical point of the equation is a zero $y^*$ of $f.$ The constant function $y=y^*$ is an equilibrium solution. Autonomous equations have the following properties:

• by uniqueness, no other solution reaches an equilibrium point;
• between consecutive equilibrium points every solution is strictly monotonic, increasing where $f>0$ and decreasing where $f<0;$
• every solution converges to an equilibrium point or diverges as $x\rightarrow\infty.$

Let $y'=f(y)$ be a first-order autonomous equation. The phase line of the equation is the record of its equilibrium points and the sign of $f$ between them, which determines the direction of every solution.

Let $y'=f(y)$ be a first-order autonomous equation and $y^*$ be an equilibrium point. The equilibrium point $y^*$ is stable if and only if every solution that starts sufficiently close to $y^*$ stays close to it for all $x\geq x_0.$

$$\forall\varepsilon>0\,\exists\delta>0:\lvert y\left(x_0\right)-y^*\rvert<\delta\Rightarrow\lvert y(x)-y^*\rvert<\varepsilon\,\,\forall\,x\geq x_0$$

The equilibrium point $y^*$ is asymptotically stable if and only if it is stable and every solution that starts sufficiently close to $y^*$ converges to it as $x\rightarrow\infty.$ The equilibrium point $y^*$ is unstable if and only if it is not stable. On the phase line, an equilibrium point is asymptotically stable if $f$ changes sign from positive to negative at it and unstable if $f$ changes sign from negative to positive.

Let $y'=f(y)$ be a first-order autonomous equation and $y^*$ be an equilibrium point. The linearisation of the equation at $y^*$ is the linear equation $u'=f'\left(y^*\right)u$ for the deviation $u=y-y^*,$ whose solutions are $u=C\,e^{f'\left(y^*\right)x}.$

Let $y'=f(y)$ be a first-order autonomous equation and $y^*$ be an equilibrium point. The linearisation criterion states that $y^*$ is asymptotically stable if $f'\left(y^*\right)<0$ and unstable if $f'\left(y^*\right)>0.$ If $f'\left(y^*\right)=0$ the criterion is inconclusive: $y'=-y^3$ has an asymptotically stable equilibrium at $0$ and $y'=y^3$ has an unstable one.

Let $y'=f(y)$ be a first-order autonomous equation and $y^*$ be an equilibrium point. The equilibrium point $y^*$ is semi-stable if and only if it attracts solutions from one side and repels them from the other, such as $0$ for $y'=y^2.$

Let $r>0$ and $K>0.$ The logistic equation is the autonomous equation

$$y'=r\,y\left(1-\frac{y}{K}\right)$$

with solution $y(x)=\frac{K}{1+\left(\frac{K}{y_0}-1\right)e^{-rx}}$ for the initial value $y(0)=y_0.$ The number $r$ is the growth rate and $K$ is the carrying capacity. The logistic equation has the following properties:

• its equilibrium points are $0,$ which is unstable since $f'(0)=r>0,$ and $K,$ which is asymptotically stable since $f'(K)=-r<0;$
• every solution with $y_0>0$ converges to $K.$