Quick memo on five basic types of first-order differential equation

When confronted with a first-order differential equation, try to convert it to one of these five basic types:

1). Separable: This is when $\frac{dy}{dx} = F(x, y)$ can be written $\frac{dy}{dx} = f(x)g(y)$ . If there is an initial condition $y(0) = A$ , we can rearrange to get an equation involving two integrals:

$\int_A^y \frac{dv}{g(v)} = \int_a^x du f(u)$

This incorporates the initial condition. It can sometimes be solved to give a solution in the form $y = h(x)$ . Note the following integration tricks which are sometimes useful here:

$\int \frac{dy}{\sqrt{1+y}} = 2\sqrt{1+y} + C$

$\int \frac{x dx}{1 + x} = \int dx \bigg(\frac{1+x}{1+x} - \frac{1}{1+x}\bigg) = \int dx - \int \frac{dx}{1+x}$

2). Homogeneous: These equations are those for which $F(x, y)$ depends only the ratio $\frac{y}{x}$ rather than on $x$ and $y$ separately. So, they are of the form $\frac{dy}{dx} = F\big(\frac{y}{x}\big)$ with initial condition $y(a) = A$ . To transform this into a separable equation, make the change of variable $y = x v$ . Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$ , so we have $F(v) = v + x \frac{dv}{dx}$ and therefore $\frac{dv}{dx} = \frac{F(v) - v}{x}$ . The initial condition becomes $A = a v(a)$ , so $v(a) = \frac{A}{a}$ . The method of separation of variables can now be applied to the transformed equation.

3). Non-separable linear first-order equations: These have the general form $\frac{dy}{dx} + y P(x) = Q(x)$ , with boundary condition $y(a) = A$ . Here, we need to find an integrating factor $p(x)$ that allows us to write the differential equation as

$\frac{d}{dx}\big(y p(x)\big) = Q(x) p(x)$

Expanding this form, we find that

$\frac{dy}{dx} p(x) + y p^{\prime}(x) = Q(x) p(x)$

and so

$\frac{dy}{dx} + y \frac{p^{\prime}(x)}{p(x)} = Q(x)$

This shows that in the original equation we must have $\frac{p^{\prime}(x)}{p(x)} = P(x)$ and therefore the required integrating factor is $p(x) = \exp\big(\int dx P(x)\big)$ . So what we do is find the integrating factor, then write the equation in the form $\frac{d}{dx}\big(y p(x)\big) = Q(x) p(x)$ . This can then be solved by integration to get $y(x) p(x) = C + \int dx \ p(x) Q(x)$ where $C$ is a constant of integration. If the integral constant cannot be evaluated directly, it is best to give the solution as

$y(x) p(x) = A + \int_a^x dt Q(t) p(t)$

$p(t) = \exp \bigg(\int_a^t du P(u) \bigg)$

because this expression automatically satisfies the boundary condition (when $x = a$ , the integral $\int_a^x dt Q(t) p(t)$ vanishes and $p(a) = \exp \big(\int_a^a du P(u) \big) = \exp(0) = 1$ so the equation reduces to $y(a) = A$ ).

4). Bernoulli’s equation: This is a nonlinear first-order equation of the form $\frac{dy}{dx} + y P(x) = y^n Q(x)$ . To solve this, make the change of variable $z = y^{1-n}$ . Then $\frac{dz}{dx} = \frac{1-n}{y^n} \frac{dy}{dx}$ , so the original equation can be rewritten as

$\frac{dz}{dx} + (1 - n) z P(x) = (1 - n) Q(x)$

This is now a linear first-order equation of the type 3) above.

5). Riccati’s equation: This is a nonlinear first-order equation of the form $\frac{dy}{dx} = P(x) + y Q(x) + y^2 R(x)$ . It reduces to a linear equation if $R(x) = 0$ and Bernoulli’s equation if $P(x) = 0$ . Otherwise, it can be reduced to a linear second-order equation by defining a new dependent variable $u(x)$ with the equation $y = - \frac{1}{uR} \frac{du}{dx}$ . This converts the original Riccati equation into