Classifying and solving first-order equations
First-order equations involve only the first derivative. A quick classification tells you which solution method will work.
What 'first order' means
A first-order ordinary differential equation relates \(y\), \(y'\), and \(x\) (or \(t\)), and can usually be written \(y' = f(x, y)\). Its general solution contains one arbitrary constant, fixed by an initial condition \(y(x_0) = y_0\). The goal is to find which tool turns the equation into ordinary integration.
Separable equations
If \(y' = g(x)h(y)\), separate and integrate: \(\int \frac{dy}{h(y)} = \int g(x)\,dx\). Example: \(y' = -2xy\) gives \(\ln|y| = -x^2 + C\), so \(y = Ae^{-x^2}\), a bell-shaped family. Remember constant solutions where \(h(y) = 0\).
Linear equations
For \(y' + P(x)y = Q(x)\), multiply by \(\mu = e^{\int P\,dx}\), making the left side \((\mu y)'\). Example: \(y' + \frac{2}{x}y = x\) for \(x > 0\) has \(\mu = x^2\), so \((x^2y)' = x^3\), giving \(y = \frac{x^2}{4} + \frac{C}{x^2}\).
\mu(x) = e^{\int P(x)\,dx}Exact equations
An equation \(M(x, y)\,dx + N(x, y)\,dy = 0\) is exact if \(M_y = N_x\). Then there is a function \(F\) with \(F_x = M\) and \(F_y = N\), and the solution is \(F(x, y) = C\). For \((2xy + 3)\,dx + (x^2 - 1)\,dy = 0\): \(M_y = 2x = N_x\), and \(F = x^2y + 3x - y\), so solutions are \(x^2y + 3x - y = C\).
M_y = N_x \Rightarrow F(x,y) = CSubstitutions
Some equations become separable or linear after a substitution. Homogeneous equations \(y' = f(y/x)\) use \(v = y/x\). Bernoulli equations \(y' + Py = Qy^n\) become linear with \(v = y^{1-n}\). Recognising these patterns expands the set of equations you can solve by hand.
Existence and uniqueness
If \(f\) and \(\partial f/\partial y\) are continuous near \((x_0, y_0)\), the initial value problem \(y' = f(x, y)\), \(y(x_0) = y_0\) has exactly one solution near \(x_0\). When the conditions fail, strange things happen: \(y' = 3y^{2/3}\), \(y(0) = 0\) has both \(y = 0\) and \(y = x^3\) as solutions.
Autonomous equations and phase lines
An autonomous equation \(y' = f(y)\) has no explicit \(x\). Its equilibria are zeros of \(f\). Drawing arrows on a number line (up where \(f > 0\), down where \(f < 0\)) shows stability without solving. For \(y' = y(1 - y)\), \(y = 0\) is unstable and \(y = 1\) is stable.
Choosing a method
Check in this order: is it separable? linear? exact? Does a substitution help? If none apply, use qualitative methods (slope fields, phase lines) or numerical methods. Most textbook equations fit one of the first three types.
Key takeaways
First-order equations have one constant. Separable equations integrate after separating; linear equations use an integrating factor; exact equations integrate to a potential \(F(x, y) = C\). Substitutions handle homogeneous and Bernoulli equations. Uniqueness holds when \(f\) and \(f_y\) are continuous, and phase lines reveal stability.
Flashcard
y' = f(x, y)
Check your understanding
Is (3x^2 + y) dx + (x + 2y) dy = 0 exact?