Ordinary Differential Equations 5 | Solve First-Order Autonomous Equations [dark version]

TL;DR

For ẋ = V(x) with continuous V, an initial value problem x(0)=x0 can be solved locally by separating variables when V(x0) ≠ 0.

Briefing Cornell Notes

Briefing

First-order autonomous differential equations admit a general, local solving method: convert the ODE into an integral involving 1/V(x), then invert an antiderivative to recover x(t). For an autonomous ODE of the form ẋ = V(x) with a continuous V on ℝ and an initial value x(0)=x0, the key move is to separate variables when V(x0) ≠ 0. In that case, near x0 the expression 1/V(x) is well-defined, so the equation can be rewritten as ẋ/V(x)=1. Integrating from time 0 to time t yields an implicit relation between x(t) and t: if F is any antiderivative of 1/V(x), then F(x(t)) − F(x0) = t (up to the constant absorbed into the antiderivative choice). Solving for x(t) requires inverting F locally, giving x(t) = F^{-1}(t + C) with the constant C fixed by the initial condition. The practical takeaway is that once an antiderivative of 1/V(x) is available, the differential equation is “almost solved”; the remaining work is algebra plus a local inverse.

Two worked examples show how the method plays out in familiar forms. For ẋ = Λx with Λ>0 and x0 ≠ 0, separation leads to (dx/x) = Λ dt. Integrating gives ln|x| = Λt + C, where the absolute value reflects the logarithm of |x| and the constant C accounts for the antiderivative’s ambiguity. Exponentiating both sides produces |x| = e^{C}e^{Λt}, and using x(0)=x0 fixes e^{C} = |x0|. The result is the standard exponential solution x(t)=x0 e^{Λt}, with the sign of x0 preserved; the absolute value disappears once the initial condition selects the correct branch.

For ẋ = x^2 with x0 ≠ 0, separation gives dx/x^2 = dt. Integrating yields −1/x = t + C, so the general implicit solution is −1/x(t) = t + C. Applying x(0)=x0 gives −1/x0 = C, and substituting back gives −1/x(t) = t − 1/x0. Solving for x(t) yields x(t) = x0 / (1 − x0 t), an explicit rational form that reveals how solutions can blow up when the denominator hits zero.

Overall, the central insight is procedural and structural: autonomous first-order equations reduce to an antiderivative-and-inversion workflow. The method is local (it relies on V(x0) ≠ 0 so 1/V(x) is integrable near x0), but it produces explicit solutions whenever the antiderivative can be found and inverted. The examples also highlight why constants matter: antiderivatives differ by additive constants, and the initial condition is what selects the correct constant to produce the specific solution x(t) that matches x0.

Cornell Notes

For an autonomous first-order ODE ẋ = V(x) with x(0)=x0, the solution can be found locally when V(x0) ≠ 0. Separating variables gives ẋ/V(x)=1, and integrating from 0 to t produces an equation in terms of an antiderivative F of 1/V(x): F(x(t)) − F(x0) = t. The remaining step is to invert F locally to get x(t) = F^{-1}(t + C), with C chosen so x(0)=x0. Two examples illustrate the workflow: ẋ=Λx leads to ln|x|=Λt+C and hence x(t)=x0 e^{Λt}; ẋ=x^2 leads to −1/x=t+C and hence x(t)=x0/(1−x0 t).

Why does the method require V(x0) ≠ 0?

When V(x0) ≠ 0, the expression 1/V(x) is finite in a neighborhood of x0 because V is continuous. That makes the separated form ẋ/V(x)=1 meaningful near the initial state, so integrating 1/V(x) is legitimate. If V(x0)=0, zeros of V can make 1/V(x) blow up or become undefined, breaking the separation/integration step at the initial point.

How does the antiderivative F of 1/V(x) determine the solution?

After separation and integration, the time variable t appears on one side and the state variable x(t) appears inside F. Specifically, with F'(x)=1/V(x), integrating gives F(x(t)) − F(x0) = t. This is an implicit solution; solving for x(t) requires a local inverse of F, yielding x(t)=F^{-1}(t+ C), where the constant is fixed by x(0)=x0.

In ẋ = Λx, where do the absolute value and the constant come from?

Separation gives dx/x = Λ dt, and integrating produces ln|x| = Λt + C. The absolute value appears because the logarithm is defined for positive arguments, so ln(x) becomes ln|x|. The constant C reflects the fact that antiderivatives of 1/x differ by additive constants; applying x(0)=x0 selects the correct C so the solution matches the initial sign and magnitude.

How does ẋ = x^2 produce a rational solution and potential blow-up?

Separation gives dx/x^2 = dt, and integrating yields −1/x = t + C. Using x(0)=x0 gives C = −1/x0, so −1/x(t)=t−1/x0. Solving gives x(t)=x0/(1−x0 t). The denominator shows where the solution becomes unbounded: when 1−x0 t = 0.

What role does the initial condition play beyond picking a constant?

The initial condition x(0)=x0 determines the additive constant that arises from integrating 1/V(x). Without that constant, the result would describe a family of solutions. With it, the implicit relation F(x(t)) − F(x0)=t becomes the specific trajectory that starts at x0 at time 0.

Review Questions

For ẋ = V(x) with x(0)=x0 and V(x0) ≠ 0, what equation involving an antiderivative F of 1/V(x) must x(t) satisfy?
In the example ẋ = Λx, why does the solution keep the sign of x0 even though ln|x| appears during integration?
In the example ẋ = x^2, at what time does the solution x(t)=x0/(1−x0 t) become singular, and how is that time determined by x0?

Key Points

1
For ẋ = V(x) with continuous V, an initial value problem x(0)=x0 can be solved locally by separating variables when V(x0) ≠ 0.
2
Separating gives ẋ/V(x)=1, and integrating yields F(x(t)) − F(x0) = t, where F'(x)=1/V(x).
3
Finding x(t) requires inverting F locally: x(t)=F^{-1}(t + C), with C fixed by the initial condition.
4
In ẋ = Λx (Λ>0), integration leads to ln|x|=Λt+C and the initial condition produces x(t)=x0 e^{Λt}.
5
In ẋ = x^2, integration leads to −1/x=t+C and the initial condition produces x(t)=x0/(1−x0 t).
6
Additive constants from antiderivatives are not optional; they are determined by enforcing x(0)=x0.
7
The method is local because it depends on 1/V(x) being well-behaved near x0, which fails if V(x0)=0.

Highlights

A universal local workflow for ẋ=V(x): integrate 1/V(x) to get F(x(t))−F(x0)=t, then invert F to obtain x(t).

For ẋ=Λx, the absolute value in ln|x| disappears after applying x(0)=x0, yielding x(t)=x0 e^{Λt}.

For ẋ=x^2, the solution x(t)=x0/(1−x0 t) makes blow-up times explicit through the denominator 1−x0 t.

The constant introduced by antiderivatives is fixed entirely by the initial condition, turning a family of solutions into one trajectory.

Topics

Autonomous ODEs
Initial Value Problems
Variable Separation
Antiderivatives and Inversion
Example Solutions