First-Order Differential Equation¶

Definition¶

A first-order differential equation is a differential equation in which only the unknown function $x(t)$ and its first derivative $x'(t)=\dfrac{\mathrm{d}x}{\mathrm{d}t}(t)$ occur. In explicit form,

x'(t)=f\big(t,x(t)\big)

for a given function $f$ of two variables. Three standard classes:

linear: $x'(t)+p(t)x(t)=q(t)$ $q (t)$ , with $p,q$ given functions of $t$ ; with constant coefficients this is $x'(t)=ax(t)+b$ $b$ ;
autonomous: $x'(t)=f\big(x(t)\big)$ $f (x (t))$ , where the right-hand side does not depend on $t$ ;
separable: $x'(t)=g(t)\,h\big(x(t)\big)$ $g (t) h (x (t))$ .

An initial value problem adds one condition $x(t_0)=x_0$ $x_{0}$ .

Interpretation¶

The equation prescribes the slope of the solution curve at every point $(t,x)$ . Starting from $x_0$ at time $t_0$ , the solution follows these slopes; one initial value is enough because the equation has order 1.

Slope segments of x' = 2tx in the (t,x) plane with solution curves Ce^(t^2), the curve e^(t^2) through x(0) = 1 in blue and the constant solution x = 0 dashed in red.

The figure draws the separable example $x'(t)=2t\,x(t)$ $2 t x (t)$ . Each grey segment has the slope $f(t,x)=2tx$ prescribed at its point; it depends on both $t$ and $x$ , flat on the line $t=0$ and steeper further out. The blue curve is the solution $\mathrm{e}^{t^2}$ fixed by the single initial value $x(0)=1$ $1$ . The grey curves are other solutions $C\mathrm{e}^{t^2}$ (here $C=2$ $2$ , $0.5$ , $-0.5$ ). The dashed line is the constant solution $x\equiv0$ $0$ , which division by $x$ loses.

Move $x(0)$ to pick one curve from the family $x(t)=3+Ce^{-2t}$ of $x'=-2x+6$ $6$ . Then change $a$ and $b$ . For $a<0$ every solution approaches the equilibrium $-b/a$ ; for $a>0$ solutions move away from it.

The constant-coefficient case $x'=ax$ is the continuous-time analogue of the first-order linear difference equation $x_{t+1}=ax_t$ $a x_{t}$ : the factor $a^t$ there becomes $\mathrm{e}^{at}$ here. The sign of $a$ , not whether $|a|<1$ $1$ , decides growth or decay.

A frequent error is to drop the constant solutions of a separable equation. Dividing by $h(x)$ is only allowed where $h(x)\neq0$ $0$ ; each zero $c$ of $h$ gives the constant solution $x(t)\equiv c$ $c$ .

Key Results¶

$x'(t)=ax(t)$ has general solution $x(t)=C\mathrm{e}^{at}$ $C e^{a t}$ ; the initial value problem with $x(t_0)=x_0$ has the unique solution $x(t)=x_0\mathrm{e}^{a(t-t_0)}$ $x_{0} e^{a (t - t_{0})}$ . This uses the exponential function and $\frac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{at}=a\mathrm{e}^{at}$ $a e^{a t}$ .
For $a\neq0$ $0$ , $x'(t)=ax(t)+b$ has general solution $x(t)=-\dfrac{b}{a}+C\mathrm{e}^{at}$ $C e^{a t}$ ; the constant $-b/a$ is its equilibrium.
If $p,q$ are continuous on an interval $I$ , the linear equation $x'+px=q$ has a unique solution on all of $I$ for each $x(t_0)=x_0$ $x_{0}$ ; it is found with an integrating factor.
Separable equations are solved by $\int \frac{\mathrm{d}x}{h(x)}=\int g(t)\,\mathrm{d}t$ wherever $h(x)\neq0$ $0$ ; the chain rule justifies this step.

Examples¶

Continuous compounding. A balance with interest rate $r=0.05$ per year satisfies $B'(t)=0.05\,B(t)$ $0.05 B (t)$ . With $B(0)=1000$ $1000$ , $B(t)=1000\mathrm{e}^{0.05t}$ $1000 e^{0.05 t}$ , so $B(10)=1000\mathrm{e}^{0.5}\approx1648.72$ $1648.72$ .

Constant forcing. For $x'(t)=-2x(t)+6$ with $x(0)=1$ $1$ : here $a=-2$ $- 2$ , $b=6$ $6$ , so $x(t)=3+C\mathrm{e}^{-2t}$ and $C=1-3=-2$ $- 2$ . Hence $x(t)=3-2\mathrm{e}^{-2t}$ $2 e^{- 2 t}$ . Check: $x'(t)=4\mathrm{e}^{-2t}$ and $-2x(t)+6=-6+4\mathrm{e}^{-2t}+6=4\mathrm{e}^{-2t}$ $4 e^{- 2 t}$ . The solution tends to the equilibrium $3$ .

Separable. For $x'(t)=2t\,x(t)$ with $x(0)=1$ $1$ : for $x>0$ $0$ , $\int\frac{\mathrm{d}x}{x}=\int2t\,\mathrm{d}t$ gives $\ln x=t^2+c$ $c$ , so $x(t)=\mathrm{e}^{t^2}$ $e^{t^{2}}$ . Check: $x'(t)=2t\mathrm{e}^{t^2}=2t\,x(t)$ $2 t x (t)$ . The function $x\equiv0$ is the constant solution excluded by the division.

Prerequisites¶

Sources¶

Math V: Advanced Linear Algebra · Week 3

Aliases: First Order Differential Equation, first-order ODE, first order ODE Keywords: x'(t)=f(t,x(t)), x'=ax+b, linear first-order equation, separable equation, continuous compounding Symbols: integral

Last updated: 14 October 2026 08:21