First order differential equations: how to solve them

Three shapes to spot first

A first order differential equation relates an unknown function and its first derivative. How you solve it depends on the form you manage to rewrite it in, and three structures keep coming back in introductory courses: the separable equation, the linear one and the homogeneous one. Recognising the structure comes before any calculation, because each one has a procedure of its own. For the general definition of a differential equation and the wider picture, there is the Recap on what differential equations are and how they work.

Separable equations

According to Paul’s Online Math Notes, an equation is separable when it can be rewritten with all the y terms, multiplied by the derivative, on one side and all the x terms on the other. The procedure then takes two moves: you separate the variables, and you integrate each side with respect to the variable that belongs to it. The integrals are the ones from calculus, and the integration is where an arbitrary constant comes from. OpenStax flags one further check: where the y factor vanishes, constant solutions appear, and they have to be examined on their own. Wolfram MathWorld gives the reason behind that check — the separation step divides by that factor, and dividing can drop exactly those solutions along the way.

Practical example: Paul’s Online Math Notes works through the initial value problem y′ = 6y²x with the condition y(1) = 1/25. Separating the variables and integrating gives the implicit solution −1/y = 3x² + c. The initial condition fixes the constant at c = −28, and the explicit solution becomes y(x) = 1/(28 − 3x²). That function holds for −√(28/3) < x < √(28/3), because the denominator vanishes at the ends of the interval.

That last step is part of the answer. The solution of an initial value problem arrives together with the interval it exists on, and the formula means something only inside it.

OpenStax adds a second case, y′ = (x² − 4)(3y + 2), which under the same method gives the general solution y = (−2 + c·e^(x³−12x))/3, with c an arbitrary constant. The plane where you read off those solution curves has its own Recap in analytic geometry.

Linear equations and the integrating factor

The standard form of a first order linear equation is y′ + p(t)y = g(t), with p and g continuous functions. OpenStax starts from an equivalent way of writing it, a(x)y′ + b(x)y = c(x), which turns into the standard one once you divide by a(x). At that point an auxiliary function comes in, the integrating factor µ: multiply both sides by µ and the derivative side becomes integrable in a single step. The factor comes out as the exponential of the integral of p, and the constant that integral brings along can be taken equal to 1. Anyone who wants to revisit linear equations in the algebraic sense can start from linear algebra.

Practical example: OpenStax applies the method to x·y′ + 3y = 4x² − 3x for x > 0. Dividing by x gives the standard form, with p(x) = 3/x and q(x) = 4x − 3, and the integrating factor computed for this equation is µ(x) = x³. Multiplying both sides by x³ and integrating leads to the general solution y = 4x²/5 − 3x/4 + c/x³. If the problem comes with an initial condition, the last step uses it to fix c.

The method extends beyond textbook exercises. Paul Dawkins applies it to a falling-body model with air resistance, where the integrating factor leads to the general solution v = 50 + c·e^(−0.196t), later fixed by the initial velocity.

Homogeneous equations and the substitution v = y/x

A first order equation is called homogeneous when it can be written as y′ = F(y/x), that is, when the right-hand side depends only on the ratio between y and x. Here you change variable: setting v = y/x makes the equation separable in the variables v and x, and you are back in the procedure above. Once the equation in v is solved, you put y/x back in place of v to recover the function you were after. Wolfram MathWorld records the same definition and confirms that the change of variable leads to a closed-form solution; in the example with an initial condition worked through by Paul Dawkins, once v goes back to y, the solution comes out explicit, and the initial condition decides which sign of the square root to keep.

How many solutions an initial value problem has

For linear equations the answer is clean: if p(t) and g(t) are continuous on an interval, the problem with an initial condition has one and only one solution on that interval. Step outside the linear case and the guarantee weakens. Uniqueness holds only locally, on an interval that may be small, and on condition that the function f and its partial derivative with respect to y be continuous near the starting point. MIT OpenCourseWare presents this existence and uniqueness theorem as a central technical result in the theory of differential equations, and points to its standard proof, Picard iteration, which builds the solution by successive approximations.

When those hypotheses fail, there can be more than one solution. Paul’s Online Math Notes works through a problem where the derivative of f with respect to y is discontinuous at y = 0 and which accepts both y = 0 and y(t) = ±(2t/3)^(3/2).